## Problems of Differentiation of Trigonometric Functions with Solutions

This post will provide you a lot of problems in the matter of derivative of trigonometric functions. You are urged to recall about the fundamental concept of trigonometry as well as the little algebra since they play a major role in this case. The problems are collected in various mathematics literatures and they also come in different difficulties. But happy news come to you. Every problem appears along with the solution. This makes you have the bigger opportunies to learn this lesson.

### Quote by Confucius

If your plan is for 1 year, plant rice; if your plan is for 10 years, plant trees; if your plan is for 100 years, educate children.

Before you start, you need to at least remember about this following fundamental derivative of trigonometric functions.

### The Derivative of Trigonometric Functions

Suppose $f(x)$ expresses a function and $f'(x)$ expresses its first derivative.
\begin{aligned} & 1.~\text{If}~f(x) = \sin x,~\text{then}~f'(x) = \cos x \\ & 2.~\text{If}~f(x) = \cos x,~\text{then}~f'(x) = -\sin x \\ & 3.~\text{If}~f(x) = \tan x,~\text{then}~f'(x) = \sec^2 x \\ & 4.~\text{If}~f(x) = \csc x,~\text{then}~f'(x) = -\cot x \csc x \\ & 5.~\text{If}~f(x) = \sec x,~\text{then}~f'(x) = \tan x \sec x \\ & 6.~\text{If}~f(x) = \cot x,~\text{then}~f'(x) = -\csc^2 x \end{aligned}Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.

Multiple Choice Section

Problem Number 1
The first derivative of $y = 3 \sin x-\cos x$ is $\cdots \cdot$
A. $3 \cos x-\sin x$
B. $3 \cos x+\sin x$
C. $\cos x-\sin x$
D. $\cos x+\sin x$
E. $5 \cos x-\sin x$

Solution

Recall that:
\begin{aligned} f(x) & = \sin x \implies f'(x) = \cos x \\ f(x) & = \cos x \implies f'(x) = -\sin x \end{aligned}
By using the facts above, we have
\begin{aligned} y & = \color{red}{3 \sin x}-\color{blue}{\cos x} \\ \implies y’ & = \color{red}{3 \cos x}-\color{blue}{(-\sin x)} \\ & = 3 \cos x + \sin x \end{aligned}
Thus, the first derivative of $y = 3 \sin x-\cos x$ is $\boxed{3 \cos x+\sin x}$

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Problem Number 2
If $g(x) = 3x^2-\dfrac{1}{2x^2}+2 \cos x$, then $g'(x)$ equals $\cdots \cdot$
A. $6x+\dfrac{1}{x^3}-2 \sin x$
B. $6x-\dfrac{1}{x^3}-2 \sin x$
C. $6x-\dfrac{1}{4x}-2 \sin x$
D. $6x+\dfrac{4}{x^3}+2 \sin x$
E. $6x+\dfrac{1}{x^3}+2 \sin x$

Solution

Recall that:
$f(x) = \cos x \implies f'(x) = -\sin x$
By using the facts above and the rule of derivative in algebra, we have
\begin{aligned} g(x) & = 3x^2-\dfrac{1}{2x^2}+2 \cos x \\ & = 3x^2-\dfrac12x^{-2}+2 \cos x \\ g'(x) & = 3(2)x-\dfrac12(-2)x^{-3}+2(-\sin x) \\ & = 6x+\dfrac{1}{x^3}-2 \sin x \end{aligned}
Thus, $\boxed{g'(x) = 6x+\dfrac{1}{x^3}-2 \sin x}$

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Problem Number 3
If $h(x) = 2 \sin x + \cos x$ ($x$ is in radian), then the value of $h’\left(\dfrac12\pi\right)$ is $\cdots \cdot$
A. $-2$                   C. $0$                  E. $2$
B. $-1$                   D. $1$

Solution

Recall that:
\begin{aligned} f(x) & = \sin x \implies f'(x) = \cos x \\ f(x) & = \cos x \implies f'(x) = -\sin x \end{aligned}
By using the facts above, we have
\begin{aligned} h(x) & = \color{red}{\sin x} + \color{blue}{\cos x} \\ \implies h'(x) & = 2 \color{red}{\cos x} + \color{blue}{(-\sin x)} \\ & = 2 \cos x-\sin x \end{aligned}
Take $x = \dfrac12\pi$.
\begin{aligned} h’\left(\dfrac12\pi\right) & = 2 \cos \dfrac12\pi-\sin \dfrac12\pi \\ & = 2(0)-1 = -1 \end{aligned}
Thus, the value of $\boxed{h’\left(\dfrac12\pi\right) = -1}$

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Problem Number 4
The differentiation of $T(x) = (\sin x + 1)(\sin x-2)$ is given by $\cdots \cdot$
A. $\sin 2x + \cos x$
B. $\sin 2x-\sin x$
C. $\sin 2x-\cos x$
D. $\cos 2x+\cos x$
E. $\cos 2x-\cos x$

Solution

Apply the product rule.
$f(x) = uv \implies f'(x) = u’v+uv’$
Given $T(x) = (\sin x + 1)(\sin x-2)$.
Let:
\begin{aligned} u & = \sin x + 1 \implies u’ = \cos x \\ v & = \sin x-2 \implies v’ = \cos x \end{aligned}
Hence, we have
\begin{aligned} T'(x) & = u’v+uv’ \\ & = (\cos x)(\sin x-2)+(\sin x+1)(\cos x) \\ & = \color{red}{\cos x \sin x}\color{blue}{-2 \cos x} \color{red}{+ \cos x \sin x} \color{blue}{+ \cos x} \\ & = 2 \sin x \cos x-\cos x \\ & = \sin 2x-\cos x \end{aligned}Note: $\boxed{\sin 2x = 2 \sin x \cos x}$
Thus, the differentiation of the function given is $\boxed{T'(x) =\sin 2x-\cos x}$

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Problem Number 5
If $h(\theta) = \left(\theta + \dfrac{\pi}{2}\right) \sin \theta$, then $h'(\theta)$ equals $\cdots \cdot$
A. $-\sin \theta-\theta \cos \theta + \dfrac{\pi}{2} \cos \theta$
B. $-\sin \theta-\theta \cos \theta -\dfrac{\pi}{2} \cos \theta$
C. $-\sin \theta+\theta \cos \theta -\dfrac{\pi}{2} \cos \theta$
D. $-\sin \theta+\theta \cos \theta +\dfrac{\pi}{2} \cos \theta$
E. $\sin \theta+\theta \cos \theta + \dfrac{\pi}{2} \cos \theta$

Solution

Apply the product rule.
$f(x) = uv \implies f'(x) = u’v+uv’$
Given $h(\theta) = \left(\theta + \dfrac{\pi}{2}\right) \sin \theta$.
Let:
\begin{aligned} u & = \theta + \dfrac{\pi}{2} \implies u’ = 1 \\ v & = \sin \theta \implies v’= \cos \theta \end{aligned}
Hence, we have
\begin{aligned} h'(\theta) & = u’v+uv’ \\ & = 1(\sin \theta)+\left( \theta + \dfrac{\pi}{2}\right)(\cos \theta) \\ & = \sin \theta + \theta \cos \theta + \dfrac{\pi}{2} \cos \theta \end{aligned}
Thus, $\boxed{h'(\theta) =\sin \theta + \theta \cos \theta + \dfrac{\pi}{2} \cos \theta}$

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Problem Number 6
The differentiation of the function $g(\theta) = \dfrac{1-\sin \theta}{\sin \theta-3}$ is $\cdots \cdot$
A. $\dfrac{-2 \cos \theta}{(\sin \theta-3)^2}$
B. $-2 \cos \theta$
C. $-2 \sin \theta$
D. $2 \sin \theta$
E. $\dfrac{2 \cos \theta}{(\sin \theta-3)^2}$

Solution

Apply the product rule.
$f(x) = \dfrac{u}{v} \implies f'(x) = \dfrac{u’v-uv’}{v^2}$
Given $g(\theta) = \dfrac{1-\sin \theta}{\sin \theta-3}$.
Let:
\begin{aligned} u & = 1-\sin \theta \implies u’ = -\cos \theta \\ v & = \sin \theta-3 \implies v’ = \cos \theta \end{aligned}
Hence, we have
\begin{aligned} g'(\theta) & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{(-\cos \theta)(\sin \theta-3)-(1-\sin \theta)(\cos \theta)}{(\sin \theta-3)^2} \\ & = \dfrac{\cancel{-\sin \theta \cos \theta} +3 \cos \theta-\cos \theta+\cancel{\sin \theta \cos \theta}}{(\sin \theta-3)^2} \\ & = \dfrac{2 \cos \theta}{(\sin \theta-3)^2} \end{aligned}Thus, the differentiation of the given function is $\boxed{\dfrac{2 \cos \theta}{(\sin \theta-3)^2}}$

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Problem Number 7
The derivative of $R(t) = \dfrac{\sin t-\cos t}{\cos t + \sin t}$ is $\cdots \cdot$
A. $1+\sin 2t$
B. $1-\sin 2t$
C. $1+\cos 2t$
D. $\dfrac{2}{1+\sin 2t}$
E. $\dfrac{-2}{1+\sin 2t}$

Solution

Apply the quotient rule.
$f(x) = \dfrac{u}{v} \implies f'(x) = \dfrac{u’v-uv’}{v^2}$
Given $R(t) = \dfrac{\sin t-\cos t}{\cos t + \sin t}$.
Let:
\begin{aligned} u & = \sin t-\cos t \implies u’ = \cos t+\sin t \\ v & = \cos t+\sin t \implies v’ = -\sin t+\cos t \end{aligned}
Hence, we have
\begin{aligned} R'(t) & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{(\cos t + \sin t)(\cos t + \sin t)-(\sin t-\cos t)(-\sin t+\cos t)}{(\cos t + \sin t)^2} \\ & = \dfrac{(\color{red}{\cos^2 t} + \sin t \cos t + \sin t \cos t \color{red}{+ \sin^2 t})-(\color{red}{-\sin^2 t}+\sin t \cos t+\sin t \cos t\color{red}{-\cos^2 t})}{\color{red}{\cos^2 t} + 2 \sin t \cos t + \color{red}{\sin^2 t}} \\ & = \dfrac{1 + \cancel{2 \sin t \cos t}-(-1)-\cancel{2 \sin t \cos t}}{1 + 2 \sin t \cos t} \\ & = \dfrac{2}{1+ \sin 2t} \end{aligned}Note: \boxed{\begin{aligned} \sin 2t & = 2 \sin t \cos t \\ \sin^2 t + \cos^2 t & = 1 \end{aligned}}
Thus, the derivative of the function is $\boxed{R'(t) = \dfrac{2}{1+ \sin 2t}}$

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Problem Number 8
If $y = \tan x-\cot x$, then the value of $\dfrac{\text{d}y}{\text{d}x}\rvert_{x = \frac{\pi}{4}}$ is $\cdots \cdot$
A. $\dfrac14$                C. $1$                E. $4$
B. $\dfrac12$                D. $2$

Solution

Recall that:
\begin{aligned} f(x) & = \tan x \implies f'(x) = \sec^2 x \\ f(x) & = \cot x \implies f'(x) = -\csc^2 x \end{aligned}
By using the facts above, we have
\begin{aligned} y & = \color{red}{\tan x}-\color{blue}{\cot x} \\ \dfrac{\text{d}y}{\text{d}x} & = \color{red}{\sec^2 x}-(\color{blue}{-\csc^2 x}) \\ & = \sec^2 x + \csc^2 x \end{aligned}
Take $x = \dfrac{\pi}{4}$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x}\rvert_{x = \frac{\pi}{4}} & = \sec^2 \dfrac{\pi}{4} + \csc^2 \dfrac{\pi}{4} \\ & = (\sqrt2)^2+(\sqrt2)^2 \\ & = 2+2 = 4 \end{aligned}
Note: $\boxed{\sec \dfrac{\pi}{4} = \csc \dfrac{\pi}{4} = \sqrt2}$
Thus, the value of the first derivative of function $y$ when $x = \dfrac{\pi}{4}$ is $\boxed{4}$

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Problem Number 9
The derivative of $y = \sec t-\csc t$ is $\cdots \cdot$
A. $\sin^3 t + \cos^3 t$
B. $\sin^3 t-\cos^3 t$
C. $\sin^2 t \cdot \cos^2 t$
D. $\dfrac{1}{(\sin t \cos t)^2}$
E. $\dfrac{\sin^3 t + \cos^3 t}{(\sin t \cos t)^2}$

Solution

Recall that:
\begin{aligned} f(x) & = \sec t \implies f'(x) = \sec t \tan t \\ f(x) & = \csc t \implies f'(x) = -\csc t \cot t \end{aligned}
By using the facts above, we have
\begin{aligned} y & = \color{blue}{\sec t}-\color{red}{\csc t} \\ y’ & = \color{blue}{(\sec t \tan t)}-\color{red}{(-\csc t \cot t)} \\ & = \dfrac{1}{\cos t} \cdot \dfrac{\sin t}{\cos t} + \dfrac{1}{\sin t} \cdot \dfrac{\cos t}{\sin t} \\ & = \dfrac{\sin t}{\cos^2 t} + \dfrac{\cos t}{\sin^2 t} \\ & = \dfrac{\sin^3 t + \cos^3 t}{(\sin t \cos t)^2} \end{aligned}
Thus, the derivative of $y = \sec t-\csc t$ is $\boxed{\dfrac{\sin^3 t + \cos^3 t}{(\sin t \cos t)^2}}$

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Problem Number 10
If $y = x^3 \tan x$, then $y’$ equals $\cdots \cdot$
A. $x^3 \tan^2 x + 3x^2 \tan x + x^3$
B. $x^3 \tan^2 x + x^2 \tan x + 3x$
C. $x^3 \tan^2 x + 3x^2 \tan x + 3x$
D. $3x^3 \tan^2 x + x^2 \tan x + x^3$
E. $3x^3 \tan^2 x + 3x^2 \tan x + x^3$

Solution

Apply the product rule.
$f(x) = uv \implies f'(x) = u’v+uv’$
Given $y = x^3 \tan x$.
Let:
\begin{aligned} u & = x^3 \implies u’ = 3x^2 \\ v & = \tan x \implies v’ = \sec^2 x \end{aligned}
Hence, we have
\begin{aligned} y’ &= u’v+uv’ \\ & = (3x^2)(\tan x)+(x^3)(\sec^2 x) \\ & = 3x^2 \tan x + (x^3)(\tan^2 x + 1) \\ & = x^3 \tan^2 x + 3x^2 \tan x + x^3 \end{aligned}
Note: $\boxed{\sec^2 x = \tan^2 x + 1}$
Thus, $\boxed{y’ = x^3 \tan^2 x + 3x^2 \tan x + x^3}$

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Problem Number 11
If $h(x) = x^2 \cot x$, then $h’\left(\dfrac{\pi}{4}\right)$ equals $\cdots \cdot$
A. $\dfrac{\pi}{8}(4+\pi)$               D. $\dfrac{\pi}{4}(8-\pi)$
B. $\dfrac{\pi}{8}(4-\pi)$                E. $\dfrac{\pi}{4}(8+\pi)$
C. $\dfrac{\pi}{8}(\pi-4)$

Solution

Apply the product rule.
$f(x) = uv \implies f'(x) = u’v+uv’$
Given $h(x) = x^2 \cot x$.
Let:
\begin{aligned} u & = x^2 \implies u’ = 2x \\ v & = \cot x \implies v’ = -\csc^2 x \end{aligned}
Hence, we have
\begin{aligned} h'(x) & = u’v+uv’ \\ & = (2x)(\cot x)+(x^2)(-\csc^2 x) \\ & = 2x \cot x-x^2 \csc^2 x \end{aligned}
Take $x = \dfrac{\pi}{4}$.
\begin{aligned} h’\left(\dfrac{\pi}{4}\right) & = 2 \cdot \dfrac{\pi}{4} \cdot \cot \dfrac{\pi}{4}-\left(\dfrac{\pi}{4}\right)^2 \csc^2 \dfrac{\pi}{4} \\ & = \dfrac{\pi}{2} \cdot 1-\dfrac{\pi^2}{16} \cdot (\sqrt2)^2 \\ & = \dfrac{\pi}{2}-\dfrac{\pi^2}{8} \\ & = \dfrac{\pi}{8}(4-\pi) \end{aligned}
So, the value of $\boxed{h’\left(\dfrac{\pi}{4}\right) = \dfrac{\pi}{8}(4-\pi)}$

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Problem Number 12
If $g(x) = \dfrac{\cos x + 2}{\sin x}$ with $\sin x \neq 0$, then the value of $g’\left(\dfrac{\pi}{2}\right)$ is $\cdots \cdot$
A. $-2$                  C. $0$                 E. $2$
B. $-1$                  D. $1$

Solution

Apply the quotient rule.
$f(x) = \dfrac{u}{v} \implies f'(x) = \dfrac{u’v-uv’}{v^2}$
Given $g(x) = \dfrac{\cos x + 2}{\sin x}$.
Let:
\begin{aligned} u & = \cos x + 2 \implies u’ = -\sin x \\ v & = \sin x \implies v’ = \cos x \end{aligned}
Hence, we have
\begin{aligned} g'(x) & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{(-\sin x)(\sin x)-(\cos x + 2)(\cos x)}{(\sin x)^2} \\ & = \dfrac{-\sin^2 x -\cos^2 x-2 \cos x}{\sin^2 x} \\ & = \dfrac{-(\color{red}{\sin^2 x + \cos^2 x})-2 \cos x}{\sin^2 x} \\ & = \dfrac{-1-2 \cos x}{\sin^2 x} \end{aligned}Take $x = \dfrac{\pi}{2}$.
\begin{aligned} g’\left(\dfrac{\pi}{2}\right) & = \dfrac{-1-2 \cos \dfrac{\pi}{2}}{\sin^2 \dfrac{\pi}{2}} \\ & = \dfrac{-1-2(0)}{(1)^2} \\ & = \dfrac{-1-0}{1} = -1 \end{aligned}
Thus, the value of $\boxed{g’\left(\dfrac{\pi}{2}\right) = -1}$

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Problem Number 13
If $f(x) = \sin x(2+\cos x)$, then the value of $f’\left(\dfrac{\pi}{4}\right) = \cdots \cdot$
A. $2\sqrt2$                     D. $\dfrac12\sqrt2$
B. $2$                           E. $\dfrac14\sqrt2$
C. $\sqrt2$

Solution

Apply the product rule.
$f(x) = uv \implies f'(x) = u’v+uv’$
Given $f(x) = \sin x(2+\cos x)$.
Let:
\begin{aligned} u & = \sin x \implies u’ & = \cos x \\ v & = 2 + \cos x \implies v’ & = -\sin x \end{aligned}
Hence, we have
\begin{aligned} f'(x) & = u’v + uv’ \\ & = (\cos x)(2+\cos x)+(\sin x)(-\sin x) \\ & = 2 \cos x + \color{red}{\cos^2 x-\sin^2 x} \\ & = 2 \cos x + \color{red}{\cos 2x} \end{aligned}Take $x = \dfrac{\pi}{4}$.
\begin{aligned} f’\left(\dfrac{\pi}{4}\right) & = 2 \cos \dfrac{\pi}{4} + \cos \cancel{2}\left(\dfrac{\pi}{\cancelto{2}{4}}\right) \\ & = \cancel{2} \cdot \dfrac{1}{\cancel{2}}\sqrt2 + 0 = \sqrt2 \end{aligned}
Thus, the value of $\boxed{f’\left(\dfrac{\pi}{4}\right) = \sqrt2}$

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Problem Number 14
The first derivative of function $y = \cos (2x^3-x^4)$ is $\cdots \cdot$
A. $y’ = \sin (2x^3-x^4)$
B. $y’ = -\sin (2x^3-x^4)$
C. $y’ = (6x^2-4x^3) \cos (2x^3-x^4)$
D. $y’ = (6x^2-4x^3) \sin (2x^3-x^4)$
E. $y’ = -(6x^2-4x^3) \sin (2x^3-x^4)$

Solution

Given $y = \cos (2x^3-x^4)$.
Apply the chain rule.
Misalkan $u = 2x^3-x^4 \implies u’ = 6x^2-4x^3$.
Hence, we have
\begin{aligned} y & = \cos u \\ \implies y’ & = -\sin u \cdot u’ \\ & = -\sin (2x^3-x^4) \cdot (6x^2-4x^3) \\ & = -(6x^2-4x^3) \sin (2x^3-x^4) \end{aligned}
Thus, the first derivative of function $y = \cos (2x^3-x^4)$ is $\boxed{y’ = -(6x^2-4x^3) \sin (2x^3-x^4)}$

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Problem Number 15
The derivative of function $g(\theta) = \cos^3 \theta$ is $\cdots \cdot$
A. $\cos \theta \sin \theta$
B. $3 \cos^2 \theta \sin \theta$
C. $-3 \cos^2 \theta \sin \theta$
B. $3 \sin^2 \theta \cos \theta$
E. $\cos^3 \theta \sin \theta$

Solution

Given $g(\theta) = \cos^{3} \theta = (\underbrace{\cos \theta}_{u})^3$.
By applying the chain rule, we have
\begin{aligned} g'(\theta) & = \color{red}{3} \cos^2 \theta \cdot \underbrace{(-\sin \theta)}_{u’} \\ & = -3 \cos^2 \theta \sin \theta \end{aligned}
Thus, the derivative of the function is $\boxed{g'(\theta) = -3 \cos^2 \theta \sin \theta}$

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Problem Number 16
The derivative of $y = \tan (2\theta-3)$ is $\cdots \cdot$
A. $\sin^2 (2\theta-3)$
B. $\cos^2 (2\theta-3)$
C. $\sec^2 (2\theta-3)$
D. $2 \sec^2 (2\theta-3)$
E. $3 \sec^2 (2\theta-3)$

Solution

Given $y = \tan \underbrace{(2\theta-3)}_{u}$.
By applying the chain rule, we have
\begin{aligned} y’ & = \sec^2 (2\theta-3) \cdot \underbrace{2}_{u’} \\ & = 2 \sec^2 (2\theta-3) \end{aligned}
Thus, the derivative of the function is $\boxed{y’= 2 \sec^2 (2\theta-3)}$

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Problem Number 17
The first derivative of $g(x) = \dfrac{\sin 2x-\cos x}{\cos 4x}$ is $g'(x)$. The value of $g’\left(\dfrac{\pi}{4}\right) = \cdots \cdot$
A. $4$                 C. $\dfrac12\sqrt2$                    E. $-1$
B. $2$                 D. $-\dfrac12\sqrt2$

Solution

Given $g(x) = \dfrac{\sin 2x-\cos x}{\cos 4x}$.
Apply the quotient rule.
Let:
\begin{aligned} u & = \sin 2x-\cos x \\ \implies u’ & = 2 \cos 2x+\sin x \\ v & = \cos 4x \\ \implies v’ & = -4 \sin 4x \end{aligned}
Hence, we have
\begin{aligned} g'(x) & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{(2 \cos 2x + \sin x)(\cos 4x)-(\sin 2x-\cos x)(-4 \sin 4x)}{(\cos 4x)^2} \\ g’\left(\dfrac{\pi}{4}\right) & = \dfrac{\left(2 \cos \dfrac{\pi}{2} + \sin \dfrac{\pi}{4}\right)(\cos \pi)-\left(\sin \dfrac{\pi}{2}-\cos \dfrac{\pi}{4}\right)(-4 \sin \pi)}{(\cos \pi)^2} \\ & = \dfrac{\left(2 \cdot 0 + \dfrac12\sqrt2\right)(-1)-\left(1-\dfrac12\sqrt2\right)(-4 \cdot 0)}{(-1)^2} \\ & = \dfrac{-\dfrac12\sqrt2-0}{1} = -\dfrac12\sqrt2 \end{aligned}Thus, the value of $\boxed{g’\left(\dfrac{\pi}{4}\right) = -\dfrac12\sqrt2}$

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Problem Number 18
The first derivative of function $h(\theta) = \sec^4 (p\theta+q)$ with $p \neq 0$ and $p, q$ be a positive real number, is $\cdots \cdot$
A. $4p \sec (p\theta+q) \cdot h(\theta)$
B. $4p \tan (p\theta+q) \cdot h(\theta)$
C. $4p \sec^4 (p\theta+q) \cdot h(\theta)$
D. $4p \tan^4 (p\theta+q) \cdot h(\theta)$
E. $4p \sec^3 (p\theta+q) \cdot \tan(p\theta + q)$

Solution

Given $\color{red}{h(\theta) = \sec^4 (p\theta+q) = (\underbrace{\sec (p\theta + q)}_{u})^4}$.
By applying the chain rule, we have
\begin{aligned} h'(\theta) & = 4 \sec^3 (p\theta+q) \cdot \underbrace{\sec (p\theta + q) \tan (p\theta+q) \cdot p}_{u’} \\ & = 4p \tan (p\theta+q) \color{red}{\sec^4 (p\theta+q)} \\ & = 4p \tan (p\theta+q) \cdot \color{red}{h(\theta)} \end{aligned}Thus, the first derivative of the trigonometric function is $\boxed{h'(\theta) = 4p \tan (p\theta+q) \cdot h(\theta)}$

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Problem Number 19
If $y = \sin 3x-\cos 3x$, then $\dfrac{\text{d}y}{\text{d}x} \rvert_{x = 45^{\circ}}$ equals $\cdots \cdot$
A. $-2$                  C. $0$                  E. $2$
B. $-1$                  D. $1$

Solution

Given $y = \sin 3x-\cos 3x$.
By using the rule of derivative of algebraic function along the chain rule, we have
\begin{aligned} y’ & = 3 \cos 3x-(-3 \sin 3x) \\ & = 3 \cos 3x + 3 \sin 3x \end{aligned}
Take $x = 45^{\circ}$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x} \rvert_{x = 45^{\circ}} & = 3 \cos 3(45^{\circ}) + 3 \sin 3(45^{\circ}) \\ & = 3 \cos 135^{\circ} + 3 \sin 135^{\circ} \\ & = 3 \cdot \left(-\dfrac12\sqrt2\right) + 3 \cdot \dfrac12\sqrt2 = 0 \end{aligned}
Thus, the value of $\boxed{\dfrac{\text{d}y}{\text{d}x} \rvert_{x = 45^{\circ}} = 0}$

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Problem Number 20
The derivative of $f(x) = x \sin x \cos x$ is $\cdots \cdot$
A. $f'(x) = \dfrac12 \sin 2x + x \cos 2x$
B. $f'(x) = x \sin 2x + x \cos 2x$
C. $f'(x) = x \sin 2x + \dfrac12 \cos 2x$
D. $f'(x) = \dfrac12 \sin 2x-x \cos 2x$
E. $f'(x) = x \sin 2x-\cos 2x$

Solution

Given $f(x) = \underbrace{x \sin x}_{u} \underbrace{\cos x}_{v}$.
Apply the product rule as
$u’ = 1(\sin x) + x(\cos x) = \sin x + x \cos x$ and $v’ = -\sin x$.
We have
\begin{aligned} f'(x) & = u’v + uv’ \\ & = (\sin x+x \cos x)(\cos x)+(x \sin x)(-\sin x) \\ & = \sin x \cos x + x \cos^2 x-x \sin^2 x \\ & = \dfrac12(2 \sin x \cos x) + x(\cos^2 x-\sin^2 x) \\ & = \dfrac12 \sin 2x + x \cos 2x \end{aligned}Note: Recall that
$\boxed{\sin 2x = 2 \sin x \cos x}$ and $\boxed{\cos 2x = \cos^2 x-\sin^2 x}$
Thus, the derivative of $f(x)$ is $\boxed{f'(x) = \dfrac12 \sin 2x + x \cos 2x}$

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Problem Number 21
If $y = (x \cos x)^2$, then $y’ = \cdots \cdot$
A. $2x \cos^2 x-2x^2 \sin x \cos x$
B. $2x \cos^2 x+2x^2 \sin x \cos x$
C. $2x \cos x+2x^2 \sin x \cos x$
D. $\cos 2x-2x^2 \sin x \cos x$
E. $\cos^2 x+2x^2 \sin x \cos x$

Solution

Given $y = (\underbrace{x \cos x}_{a})^2$.
First, we will find the derivative of $a$ by using the product rule.
Let:
\begin{aligned} u & = x \implies u’ = 1 \\ v & = \cos x \implies v’ = -\sin x \end{aligned}
Therefore, we will have
\begin{aligned} a’ & = u’v + uv’ \\ & = 1(\cos x)+x(-\sin x) \\ & = \cos x-x \sin x \end{aligned}
Now apply the chain rule.
\begin{aligned} y’ & = 2a \cdot a’ \\ & = 2(x \cos x)(\cos x-x \sin x) \\ & = 2x \cos^2 x-2x^2 \sin x \cos x \end{aligned}
So, the derivative of the function is given by $\boxed{y’ = 2x \cos^2 x-2x^2 \sin x \cos x}$

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Problem Number 22
The differentiation of $y = \sin^3 (2x+3)$ is $\cdots \cdot$
A. $-\dfrac32(\cos (6x + 9)-\cos (2x+3))$
B. $\dfrac94(\cos (2x+3)-\cos (6x+9))$
C. $\dfrac34(\cos (2x+3)-\cos (6x+9))$
D. $\dfrac34(\cos (6x + 9)-\cos (2x+3))$
E. $\dfrac32(\cos (6x + 9)-\cos (2x+3))$

Solution

Given $y = \sin^3 (2x+3) = (\underbrace{\sin (2x+3)}_{u})^3$.
By applying the chain rule, we have
\begin{aligned} y’ & = 3 \sin^2 (2x+3) \cdot \underbrace{(2 \cos (2x + 3))}_{u’} \\ & = 6 \sin (2x+3) \sin (2x+3) \cos (2x+3) \\ & = \cancelto{3}{6} \sin (2x+3) \cdot \dfrac{1}{\cancel{2}} \sin 2(2x+3) && (\sin A \cos A = \dfrac12 \sin 2A) \\ & = 3 \sin (2x+3) \sin (4x+6) \\ & = -\dfrac32\left[\cos ((2x+3)+(4x+6))-\cos ((2x+3)-(4x+6))\right] && (\sin A \sin B = -\dfrac12[\cos (A+B)-\cos(A-B)]) \\ & = -\dfrac32(\cos (6x+9)-\cos (-2x-3)) \\ & = -\dfrac32(\cos (6x+9)-\cos (2x+3)) \end{aligned}Thus, the differentiation of $y$ is $\boxed{y’ = -\dfrac32(\cos (6x+9)-\cos (2x+3))}$

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Problem Number 23
If given $y = \sqrt{1+\sin^2 x}$, then $y’ = \cdots \cdot$
A. $\dfrac{\sin x + \cos x}{1+\sin^2 x}$
B. $\dfrac{\sin x + \cos x}{\sqrt{1+\sin^2 x}}$
C. $\dfrac{\sin x-\cos x}{1+\sin^2 x}$
D. $\dfrac{\sin x-\cos x}{\sqrt{1+\sin^2 x}}$
E. $\dfrac{\sin x \cos x}{\sqrt{1+\sin^2 x}}$

Solution

Given $y = \sqrt{1+\sin^2 x} = (1+\sin^2 x)^{\frac12}$.
By applying the chain rule, we have
\begin{aligned} y’ & = \dfrac12(1+\sin^2 x)^{-\frac12} \cdot D_x(1+\sin^2 x) \\ & = \dfrac{1}{\cancel{2}}(1+\sin^2 x)^{-\frac12} \cdot (\cancel{2} \sin x) D_x(\sin x) \\ & = (1 + \sin^2 x)^{-\frac12} \sin x \cos x \\ & = \dfrac{\sin x \cos x}{\sqrt{1+\sin^2 x}} \end{aligned}
Thus, the derivative of $y$ is $\boxed{y’ = \dfrac{\sin x \cos x}{\sqrt{1+\sin^2 x}}}$

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Problem Number 24 ($\bigstar$ HOTS $\bigstar$)
If $y = \sin(\sin(\sin(\cdots \sin(\sin x))\cdots))$, then the value of $\dfrac{\text{d}y}{\text{d}x}$ at $x=0$ is $\cdots \cdot$
A. $-\infty$                  C. $0$                   E. $\infty$
B. $-1$                    D. $1$

Solution

Observe the equation $y_1 = \sin x$.
Its first derivative is given by $y_1′ = \cos x$.
Now, observe the equation $y_2 = \sin (\underbrace{\sin x}_{u})$.
Its derivative can be evaluated by applying the chain rule as follows.
$y_2’= \underbrace{\cos x}_{u’} \cos (\sin x)$
Next, observe the equation $y_3 = \sin(\underbrace{\sin(\sin x))}_{u})$.
Its derivative also can be evaluated by applying the chain rule as follows.
$y_3’= \underbrace{\cos x \cos (\sin x)}_{u’} \cos(\sin(\sin x)))$
Looking at the pattern of its derivatives, it can be concluded that the first derivative of $y = \sin(\sin(\sin(\cdots \sin(\sin x))\cdots))$ has the expressions of $\color{red}{\cos x}$, $\color{red}{\cos(\sin x)}$, $\cdots$, and $\color{red}{\cos(\sin(\sin(\cdots \sin(\sin x))\cdots))}$.
Notice that $\sin 0 = 0$ and $\cos 0 = 1$.
Substituting $x = 0$ will lead us to
$y’_{x = 0} = \cos 0 \cdot \cos 0 \cdots \cos 0 = 1$

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Problem Number 25 ($\bigstar$ OLYMPIAD $\bigstar$)
If $f(x) = \dfrac{\sec x + \csc x}{\csc x \sec x}$, then $f^{(2019)}(x)$ (the $2019$th derivative) equals $\cdots \cdots$
A. $\sin x + \cos x$
B. $\cos x-\sin x$
C. $\sec x+\csc x$
D. $-\sin x-\cos x$
E. $-\cos x+\sin x$

Solution

Simplify $f(x)$ first as follows.
\begin{aligned} f(x) & = \dfrac{\sec x + \csc x}{\csc x \sec x} \\ & = \dfrac{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}{\dfrac{1}{\sin x \cos x}} \\ & = \dfrac{\sin x \bcancel{\cos x}}{\cancel{\cos x}} + \dfrac{\cancel{\sin x} \cos x}{\cancel{\sin x}} \\ & = \sin x + \cos x \end{aligned}
Look at the derivative patterns.
\begin{aligned} f'(x) & = \cos x-\sin x \\ f^{\prime \prime}(x) & = -\sin x-\cos x \\ f^{\prime \prime \prime}(x) & = -\cos x+\sin x \\ f^{(4)}(x) & = \sin x+\cos x \end{aligned}
Apparently, the fourth derivative of function $f(x)$ equals to itself.
This means that every $4n$th of derivatives for natural number $n$ equals to $f(x) = \sin x + \cos x$. This rule is also hold for the $(4n+1)$th derivatives, $(4n+2)$th derivatives, and $(4n+3)$th derivatives.
Since $2019$ has a remainder of $3$ after divided by $4$, then the $2019$th derivative of function $f$ equals to its third derivative, that is
$\boxed{f^{(2019})(x) = f^{\prime \prime \prime}(x) = -\cos x+\sin x}$

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Problem Number 26
The equation for the tangent line of the curve $y = \tan x$ at the point $\left(\dfrac{\pi}{4}, 1\right)$ is $\cdots \cdot$
A. $y = 2x + \left(1+\dfrac{\pi}{2}\right)$
B. $y = 2x + \left(\dfrac{\pi}{2}-1\right)$
C. $y = 2x + \left(1-\dfrac{\pi}{2}\right)$
D. $y = 2x + (2-\pi)$
E. $y = 2x + (2+\pi)$

Solution

Given $y = \tan x$ and tangent point $\left(\dfrac{\pi}{4}, 1\right)$.
First, we will find the derivative of $y$. It is simply $y’ = \sec^2 x$.
Plug $x = \dfrac{\pi}{4}$ into $y’$ so we have the slope.
$m = \sec^2 \dfrac{\pi}{4} = (\sqrt2)^2 = 2$
The equation for the tangent line that runs through $(x_1, y_1) = \left(\dfrac{\pi}{4}, 1\right)$ and a slope of $m = 2$ is
\begin{aligned} y & = m(x-x_1)+y_1 \\ & = 2\left(x-\dfrac{\pi}{4}\right)+1 \\ & = 2x-\dfrac{\pi}{2}+1 \\ & = 2x + \left(1-\dfrac{\pi}{2}\right) \end{aligned}
Thus, the equation for the tangent line is $\boxed{y = 2x + \left(1-\dfrac{\pi}{2}\right)}$
The graph is shown below.

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Problem Number 27
The tangent line that passes through the curve $y = \sin x + \cos x$ at the point having abscissa of $\dfrac{\pi}{2}$ will intersect $Y$-axis at the point whose ordinate is $\cdots \cdot$
A. $-\dfrac{\pi}{2} + 1$                      D. $\dfrac{\pi}{2}$
B. $-\dfrac{\pi}{2}$                            E. $\dfrac{\pi}{2} + 1$
C. $-\dfrac{\pi}{2}- 1$

Solution

Given $y = \sin x + \cos x$.
Plug $x = \dfrac{\pi}{2}$ to get
$y = \sin \dfrac{\pi}{2} + \cos \dfrac{\pi}{2}= 1 + 0 = 1$
The tangent point is at $\left(\dfrac{\pi}{2}, 1\right)$.
The derivative of $y$ is $y’ = \cos x-\sin x$.
The slope of the tangent line$(m)$ is represented as the value of $y’$ at $x = \dfrac{\pi}{2}$, that is
$y’ = m = \cos \dfrac{\pi}{2}-\sin \dfrac{\pi}{2} = 0-1 = -1$
The line that runs through $(x_1, y_1) = \left(\dfrac{\pi}{2}, 1\right)$ and has a gradien of $m = -1$ is
\boxed{\begin{aligned} y-y_1 & = m(x-x_1) \\ y-1 & = -1\left(x-\dfrac{\pi}{2}\right) \\ y & = -x + \dfrac{\pi}{2} + 1 \end{aligned}}
This line intersects $Y$-axis at $x = 0$ so we have
$\boxed{y = 0 + \dfrac{\pi}{2} + 1 = \dfrac{\pi}{2} + 1}$
The graph is shown below.

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Essay Section

Problem Number 1
Find $\dfrac{\text{d}y}{\text{d}x}$ for the following functions.
a. $y = 2x + \sin x$
b. $y = 5 \sin x-6 \cos x$
c. $y = 8x^3-\sin x +6$
d. $y = 6 \cos x-8(x^2+x)$
e. $y = 2 \sin x + \tan x$
f. $y = x^2 + \cos x + \cos \dfrac{\pi}{4}$

Solution

Given $y = 2x + \sin x$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = \dfrac{\text{d}}{\text{d}x}(2x) + \dfrac{\text{d}}{\text{d}x}(\sin x) \\ & = 2 + \cos x \end{aligned}
Thus, its derivative is $\boxed{\dfrac{\text{d}y}{\text{d}x} = 2 + \cos x}$
Given $y = 5 \sin x-6 \cos x$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = 5 \dfrac{\text{d}}{\text{d}x}(\sin x)-6 \dfrac{\text{d}}{\text{d}x}(\cos x) \\ & = 5 \cos x-6(-\sin x) \\ & = 5 \cos x + 6 \sin x \end{aligned}
Thus, its derivative is $\boxed{\dfrac{\text{d}y}{\text{d}x} = 5 \cos x + 6 \sin x}$
Given $y = 8x^3-\sin x + 6$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = 8 \dfrac{\text{d}}{\text{d}x}(x^3)- \dfrac{\text{d}}{\text{d}x}(\sin x) + \dfrac{\text{d}}{\text{d}x} (6) \\ & = 8(3x^2)-\cos x+0 \\ & = 24x^2-\cos x \end{aligned}
Thus, its derivative is $\boxed{\dfrac{\text{d}y}{\text{d}x} = 24x^2-\cos x}$
Given $y = 6 \cos x-8(x^2+x)$.
The equation above is equivalent to $y = 6 \cos x-8x^2-8x$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = 6 \dfrac{\text{d}}{\text{d}x}(\cos x)-8\dfrac{\text{d}}{\text{d}x}(x^2)-8\dfrac{\text{d}}{\text{d}x} (x) \\ & = 6(-\sin x)-8(2x)-8(1) \\ & = -6 \sin x-16x-8 \end{aligned}
Thus, its derivative is $\boxed{\dfrac{\text{d}y}{\text{d}x} = -6 \sin x-16x-8}$
Given $y = 2 \sin x + \tan x$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = 2 \dfrac{\text{d}}{\text{d}x}(\sin x)+\dfrac{\text{d}}{\text{d}x}(\tan x) \\ & = 2(\cos x)+\sec^2 x \\ & = 2 \cos x + \sec^2 x \end{aligned}
Thus, its derivative is $\boxed{\dfrac{\text{d}y}{\text{d}x} = 2 \cos x + \sec^2 x}$
Given $y = x^2 + \cos x + \cos \dfrac{\pi}{4}$.
\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = \dfrac{\text{d}}{\text{d}x}(x^2)+\dfrac{\text{d}}{\text{d}x}(\cos x) + \dfrac{\text{d}}{\text{d}x}\left(\cos \dfrac{\pi}{4}\right) \\ & = 2x + (-\sin x) + 0 \\ & = 2x-\sin x \end{aligned}
Thus, its derivative is $\boxed{\dfrac{\text{d}y}{\text{d}x} = 2x-\sin x}$

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Problem Number 2
Find the value of $f’\left(\dfrac{\pi}{4}\right)$ for the following functions.
a. $f(t) = 2 \sec t + 3 \tan t-\tan \dfrac{\pi}{3}$
b. $f(\theta) = 2 \sin \theta + 3 \cos \theta$
c. $f(t) = \sin^2 t + \cos^2 t$
d. $f(\alpha) = \csc \alpha + \sec \alpha$
e. $f(x) = \tan x-\cot x$

Solution

Find the function’s derivative and then plug its variable as $\dfrac{\pi}{4}$.
Given $f(t) = 2 \sec t + 3 \tan t-\tan \dfrac{\pi}{3}$.
\begin{aligned} f'(t) & = 2 (\sec t \tan t) + 3 (\sec^2 t)-0 \\ f’\left(\dfrac{\pi}{4}\right) & = 2 \cdot \sec \dfrac{\pi}{4} \cdot \tan \dfrac{\pi}{4} + 3 \sec^2 \dfrac{\pi}{4} \\ & = 2 \cdot \sqrt2 \cdot 1 + 3 (\sqrt2)^2 \\ & = 2\sqrt2 + 6 \end{aligned}
Thus, the value of $\boxed{f’\left(\dfrac{\pi}{4}\right) = 2\sqrt2+6}$
Given $f(\theta) = 2 \sin \theta + 3 \cos \theta$.
\begin{aligned} f'(\theta) & = 2 \cos \theta + 3(-\sin \theta) \\ & = 2 \cos \theta-3 \sin \theta \\ f’\left(\dfrac{\pi}{4}\right) & = 2 \cos \dfrac{\pi}{4}-3 \sin \dfrac{\pi}{4} \\ & = 2 \cdot \dfrac12\sqrt2-3 \cdot \dfrac12\sqrt2 \\ & = \sqrt2-\dfrac32\sqrt2 = -\dfrac12\sqrt2 \end{aligned}
Thus, the value of $\boxed{f’\left(\dfrac{\pi}{4}\right) = -\dfrac12\sqrt2}$
Given $f(t) = \sin^2 t + \cos^2 t$.
Notice that $f(t) = 1$ since $\sin^2 t + \cos^2 t = 1$ (Pythagorean Identity). In other words, $f(t)$ is merely a constant function.
This means, $f'(t) = 0$ since the derivative of any constant function is always zero.
Thus, $\boxed{f’\left(\dfrac{\pi}{4}\right) = 0}$
Given $f(\alpha) = \csc \alpha + \sec \alpha$.
By using the addition rule of derivative, we have
$f'(\alpha) = (-\csc \alpha \cot \alpha) + (\sec \alpha \tan \alpha)$
Plug $\alpha = \dfrac{\pi}{4}$ to get
\begin{aligned} f’\left(\dfrac{\pi}{4}\right) & = \left(-\csc \dfrac{\pi}{4} \cot \dfrac{\pi}{4}\right) + \left(\sec \dfrac{\pi}{4} \tan \dfrac{\pi}{4}\right) \\ & = (-\sqrt2 \cdot 1) + (\sqrt2 \cdot 1) \\ & = -\sqrt2 + \sqrt2 = 0 \end{aligned}Thus, $\boxed{f’\left(\dfrac{\pi}{4}\right) = 0}$
Given $f(x) = \tan x-\cot x$.
By using the subtraction rule of derivative, we have
\begin{aligned} f'(x) & = \sec^2 x-(-\csc^2 x) \\ & = \sec^2 x + \csc^2 x \end{aligned}
Plug $x = \dfrac{\pi}{4}$ to get
\begin{aligned} f’\left(\dfrac{\pi}{4}\right) & = \sec^2 \dfrac{\pi}{4} + \csc^2 \dfrac{\pi}{4} \\ & = (\sqrt2)^2 + (\sqrt2)^2 \\ & = 2+2 = 4 \end{aligned}
Thus, $\boxed{f’\left(\dfrac{\pi}{4}\right) = 4}$

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Problem Number 3
If $h(x) = x \cdot g(x)$ with
$g(x) = \dfrac{a \sin x + b \cos x}{c \sin x + d \cos x}$,
and $(c \sin x + d \sin x) \neq 0$, show that $x \cdot h'(x) = h(x) + x^2g'(x)$.

Solution

Given $h(x) = x \cdot g(x)$.
By applying the product rule, we have
$h'(x) = 1 \cdot g(x) + x \cdot g'(x)$.
Hence,
\begin{aligned} x \cdot h'(x) & = x(g(x) + x \cdot g'(x)) \\ & = x \cdot g(x) + x^2 \cdot g'(x) \\ & = h(x) + x^2 \cdot g'(x) \end{aligned}
Thus, it is shown that $\boxed{x \cdot h'(x) = h(x) + x^2g'(x)}$

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Problem Number 4
Show that the first derivative of function
$f(x) = \dfrac{a \sin x + b \cos x}{c \sin x + d \cos x}$
is
$f'(x) = \dfrac{ad-bc}{(c \sin x + d \cos x)^2}$.

Solution

Given $f(x) = \dfrac{a \sin x + b \cos x}{c \sin x + d \cos x}$.
Applying the quotient rule, first we assume
\begin{aligned} & u = a \sin x + b \cos x \implies u’ = a \cos x-b \sin x \\ & v = c \sin x + d \cos x \implies v’ = c \cos x-d \sin x \end{aligned}Therefore, we have
\begin{aligned} f'(x) & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{(a \cos x-b \sin x) (c \sin x + d \cos x)-(a \sin x + b \cos x)(c \cos x-d \sin x)}{(c \sin x + d \cos x)^2} \end{aligned}Notice that to prove the statement above, we only need to observe the numerator, since their denominators are already same. We express the numerator of f'(x) as follows.\begin{aligned} g'(x) & = (a \cos x-b \sin x) \cdot (c \sin x + d \cos x)-(a \sin x + b \cos x)(c \cos x-d \sin x) \\ & = \cancel{ac \sin x \cos x} + ad \cos^2 x-bc \sin^2 x-\bcancel{bd \sin x \cos x}-(\cancel{ac \sin x \cos x}-ad \sin^2 x + bc \cos^2 x-\bcancel{bd \sin x \cos x}) \\ & = ad \sin^2 x + ad \cos^2 x-bc \sin^2 x-bc \cos^2 x \\ & = ad(\sin^2 x + \cos^2 x)-bc(\sin^2 x + \cos^2 x) \\ & = ad(1)-bc(1) = ad-bc \end{aligned}Thus, it is showed that the first derivative of f(x) is \boxed{f'(x)= \dfrac{ad-bc}{(c \sin x + d \cos x)^2}} [collapse] Problem Number 5 By applying the following formula: f(x) = \dfrac{a \sin x + b \cos x}{c \sin x + d \cos x} whose derivative is given by f'(x) = \dfrac{ad-bc}{(c \sin x + d \cos x)^2}, evaluate the derivative of the following functions. a. g(\theta) = \dfrac{\sin \theta}{\sin \theta + \cos \theta} b. h(t) = \dfrac{2 \sin t-\cos t}{3 \sin t + \cos t} c. p(\alpha) = \dfrac{\sin \alpha + 2 \cos \alpha}{3 \sin \alpha} d. k(x) = \dfrac{4 \cos x-2 \sin x}{5 \cos x + 3 \sin x} Solution Answer a) Express the function into the general form of the given formula. g(\theta) = \dfrac{1 \sin \theta + 0 \cos \theta}{1 \sin \theta + 1 \cos \theta} We have a = c = d = 1, and b = 0. Hence, its derivative is given by \begin{aligned} g'(\theta) & = \dfrac{ad-bc}{(c \sin \theta + d \cos \theta)^2} \\ & = \dfrac{1(1)-0(1)}{(\sin \theta + \cos \theta)^2} \\ & = \dfrac{1}{(\sin \theta + \cos \theta)^2} \end{aligned} Answer b) Given h(t) = \dfrac{2 \sin t-\cos t}{3 \sin t + \cos t}. The function h is already in the form of the formula, with a = 2, b= -1, c = 3, and d = 1. The derivative of this function is \begin{aligned} h'(t) & = \dfrac{ad-bc}{(c \sin t + d \cos t)^2} \\ & = \dfrac{2(1)-(-1)(3)}{(3 \sin t + \cos t)^2} \\ & = \dfrac{5}{(3 \sin t + \cos t)^2} \end{aligned} Answer c) Express the function into the general form of the given formula. p(\alpha) = \dfrac{1 \sin \alpha + 2 \cos \alpha}{3 \sin \alpha + 0 \cos \alpha} We have a = 1, b = 2, c = 3, and d = 0. Hence, its derivative is given by \begin{aligned} p'(\alpha) & = \dfrac{ad-bc}{(c \sin \alpha + d \cos \alpha)^2} \\ & = \dfrac{1(0)-(2)(3)}{(3 \sin \alpha + 0 \cos \alpha)^2} \\ & = -\dfrac{6}{9 \sin^2 \alpha} = -\dfrac{2}{3 \sin^2 \alpha} \end{aligned} Answer d) Express the function into the general form of the given formula. k(x) = \dfrac{-2 \sin x + 4 \cos x}{3 \sin x + 5 \cos x} We have a = -2, b = 4, c = 3, and d = 5. Hence, its derivative is given by \begin{aligned} k'(x) & = \dfrac{ad-bc}{(c \sin x + d \cos x)^2} \\ & = \dfrac{(-2)(5)-(4)(3)}{(3 \sin x + 5 \cos x)^2} \\ & = -\dfrac{22}{(3 \sin x + 5 \cos x)^2} \end{aligned} [collapse] Problem Number 6 Given functions f(\theta) = \sin \theta and g(\theta) = \cos \theta. Show that: a. f'(\theta) \cdot g(\theta)-f(\theta) \cdot g'(\theta) = 1 b. f'(\theta) \cdot g(\theta)+f(\theta) \cdot g'(\theta) = 2 \cos^2 \theta-1 Solution Given: f(\theta) = \sin \theta \implies f'(\theta) = \cos \theta g(\theta) = \cos \theta \implies g'(\theta) = -\sin \theta Answer a) \begin{aligned} & f'(\theta) \cdot g(\theta)-f(\theta) \cdot g'(\theta) \\ & = \cos \theta \cdot (\cos \theta)-\sin \theta \cdot (-\sin \theta) \\ & = \cos^2 \theta + \sin^2 \theta = 1 \end{aligned} (Q.E.D) Answer b) \begin{aligned} & f'(\theta) \cdot g(\theta)+f(\theta) \cdot g'(\theta) \\ & = \cos \theta \cdot (\cos \theta)+\sin \theta \cdot (-\sin \theta) \\ & = \cos^2 \theta- \sin^2 \theta \\ & = \cos^2 \theta-(1-\cos^2 \theta) \\ & = 2 \cos^2 \theta-1 \end{aligned} (Q.E.D) [collapse] Problem Number 7 If f(x) = \dfrac{x \sin x}{\sin x + \cos x}, with (\sin x + \cos x) \neq 0, prove that f'(x)(1+ \sin 2x) = x + \sin x(\sin x + \cos x). Solution Given f(x) = \dfrac{x \sin x}{\sin x + \cos x}. We will find the first derivative of f(x) by applying the quotient rule. Let:\begin{aligned} u & = x \sin x \implies u’ = \sin x + x \cos x \\ v & = \sin x + \cos x \implies v’ = \cos x-\sin x \end{aligned}$$Hence,$$\begin{aligned} f'(x) & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{(\sin x + x \cos x)(\sin x + \cos x)-(x \sin x)(\cos x-\sin x)}{(\sin x + \cos x)^2} \\ & = \dfrac{\sin^2 x + \sin x \cos x +\cancel{ x \sin x \cos x} + x \cos^2 x- \cancel{x \sin x \cos x} + x \sin^2 x}{(\sin^2 x + \cos^2 x) + 2 \sin x \cos x} \\ & = \dfrac{\sin^2 x + \sin x \cos x + x(\sin^2 x + \cos^2 x)}{1 + \sin 2x} \\ & = \dfrac{\sin x(\sin x + \cos x) + x(1)}{1+\sin 2x} \\ & = \dfrac{x + \sin x(\sin x + \cos x)}{1+\sin 2x} \end{aligned}Next, we have \begin{aligned} & f'(x)(1+\sin 2x) \\ & = \dfrac{x + \sin x(\sin x + \cos x)}{\cancel{1+\sin 2x}} \cdot \cancel{(1+\sin 2x)} \\ & = x + \sin x(\sin x + \cos x) \end{aligned} (Q.E.D) [collapse] Problem Number 8 Evaluate the derivative of the following trigonometric function. f(t) = \dfrac{2t \sin t + 3 \cos t}{\sin t + 3t \cos t} Solution Use the quotient rule. Let: \begin{aligned} u & = 2t \sin t + 3 \cos t \\ u’ & = (2 \sin t + 2t \cos t)-3 \sin t \\ & = -\sin t + 2t \cos t \\ v & = \sin t + 3t \cos t \\ v’ & = \cos t + (3 \cos t -3t \sin t) \\ & = 4 \cos t-3t \sin t \end{aligned} Hence, we have\begin{aligned} f'(t) & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{(-\sin t + 2t \cos t)(\sin t + 3t \cos t)-(2t \sin t + 3 \cos t)(4 \cos t-3t \sin t)}{(\sin t + 3t \cos t)^2} \\ & = \dfrac{\color{red}{-\sin^2 t} -3t \sin t \cos t + 2t \sin t \cos t + \color{blue}{6t^2 \cos^2 t} + 3 \sin t \cos t + 9t \cos^2 t-8t \sin t \cos t+\color{red}{6t^2 \sin^2 t}\color{blue}{-12 \cos^2 t}+9t \sin t \cos t}{(\sin t + 3t \cos t)^2} \\ & = \dfrac{(6t^2-1) \sin^2 t + (6t^2-12) \cos^2 t + (-3t+2t-8t+9t) \sin t \cos t}{(\sin t + 3t \cos t)^2} \\ & = \dfrac{(6t^2-1) \sin^2 t + (6t^2-12) \cos^2 t}{(\sin t + 3t \cos t)^2} \end{aligned}Thus, the derivative of the function is \boxed{f'(t) = \dfrac{(6t^2-1) \sin^2 t + (6t^2-12) \cos^2 t}{(\sin t + 3t \cos t)^2}} [collapse] Problem Number 9 Given that f(x) = \dfrac{\cos x-\sin x}{\cos x + \sin x} with (\cos x + \sin x) \neq 0. Prove that f'(x) = -1(1 + f^2(x)). Solution Use the quotient rule. \boxed{f(x) = \dfrac{u}{v} \implies f'(x) = \dfrac{u’v-uv’}{v^2}} Given f(x) = \dfrac{\cos x-\sin x}{\cos x + \sin x}. Let: \begin{aligned} u & = \color{red}{\cos x} \color{blue}{-\sin x} \\ u’ & = \color{red}{-\sin x}\color{blue}{-\cos x} = -(\cos x + \sin x) \\ v & = \color{red}{\cos x}\color{blue}{+\sin x} \\ v’ & = \color{red}{-\sin x}\color{blue}{+\cos x} = \cos x-\sin x \end{aligned} Hence,\begin{aligned} f'(x) & = \dfrac{-(\cos x + \sin x)(\cos x + \sin x)-(\cos x-\sin x)(\cos x-\sin x) }{(\cos x + \sin x)^2} \\ & = -1\left(\dfrac{(\cos x + \sin x)^2 + (\cos x-\sin x)^2}{(\cos x + \sin x)^2}\right) \\ & = -1\left(\dfrac{(\cos x + \sin x)^2}{(\cos x + \sin x)^2} + \dfrac{(\cos x-\sin x)^2}{(\cos x + \sin x)^2}\right) \\ & = -1\left(1+\left(\dfrac{\cos x-\sin x}{\cos x+\sin x}\right)^2\right) \\ f'(x) & = -1(1 + f^2(x)) \end{aligned}$$(Q.E.D) [collapse] Problem Number 10 Find the derivative of the following function h(y) = y^3-y^2 \cos y + 2y \sin y + 2 \cos y Solution Given h(y) = y^3-\color{red}{y^2 \cos y} + \color{blue}{2y \sin y} + 2 \cos y. The expression in blue above is the product of two functions, so its derivative can be evaluated by applying the product rule: f(x) = uv \implies f'(x) = u’v + uv’. Hence, we have$$\begin{aligned} h'(y) & = 3y^2-(2y \cos y + y^2 (-\sin y)) + 2(\sin y + y \cos y) + 2 (-\sin y) \\ & = 3y^2-\cancel{2y \cos y}+y^2 \sin y + \bcancel{2 \sin y} + \cancel{2y \cos y}-\bcancel{2 \sin y} \\ & = 3y^2+y^2 \sin y \end{aligned}Thus, the derivative of the trigonometric function is \boxed{h'(y) = 3y^2 + y^2 \sin y} [collapse] Problem Number 11 Find the equation for the tangent line on the curve of the following trigonometric functions at the given point. 1. f(x) = \sin x at the point whose abscissa is x = \dfrac{\pi}{6}. 2. f(x) = \cot x-2 \csc x at the point whose abscissa is x = \dfrac{\pi}{3}. Solution Answer a) For x = \dfrac{\pi}{6}, we have f\left(\dfrac{\pi}{6}\right) = \sin \dfrac{\pi}{6} = \dfrac12 The tangent point is at \left(\dfrac{\pi}{6}, \dfrac12\right). The first derivative of function f(x)= \sin x is f'(x) = \cos x. The slope of the tangent line is represented as the value of function f’ at x = \dfrac{\pi}{6}, that is m = f’\left(\dfrac{\pi}{6}\right) = \cos \dfrac{\pi}{6} = \dfrac12\sqrt3. The equation for the tangent line that runs through (x_1, y_1) = \left(\dfrac{\pi}{6}, \dfrac12\right) having a slope of m = \dfrac12\sqrt3 is \begin{aligned} y & = m(x-x_1)+y_1 \\ y & = \dfrac12\sqrt3\left(x-\dfrac{\pi}{6}\right) + \dfrac12 \\ 2y & = \sqrt3\left(x-\dfrac{\pi}{6}\right) +1 \end{aligned} Thus, its tangent line is given by \boxed{2y = \sqrt3\left(x-\dfrac{\pi}{6}\right) +1} Answer b) For x = \dfrac{\pi}{3}, we have \begin{aligned} f\left(\dfrac{\pi}{3}\right) & = \cot \dfrac{\pi}{3}-2 csc \dfrac{\pi}{3} \\ & = \dfrac{\sqrt3}{3}-2 \cdot \dfrac23\sqrt3 \\ & = (1-4)\dfrac{\sqrt3}{3} = -\sqrt3 \end{aligned} The tangent point is at \left(\dfrac{\pi}{3}, -\sqrt3\right). The first derivative of function f(x)= \cot x-2 \csc x is \begin{aligned}vf'(x) & = -\csc^2 x-2(-\csc x \cot x) \\ & = 2 \csc x \cot x-\csc^2 x \end{aligned} The slope of the tangent line is represented as the value of function f’ at x = \dfrac{\pi}{3}, yaitu \begin{aligned} m & = f’\left(\dfrac{\pi}{3}\right) \\ & = 2 \csc \dfrac{\pi}{3} \cot \dfrac{\pi}{3} -\csc^2 \dfrac{\pi}{3} \\ & = 2 \cdot \dfrac23\sqrt3 \cdot \dfrac13\sqrt3-\left(\dfrac23\sqrt3\right)^2 \\ & = \dfrac43-\dfrac{4}{9}(3) = 0 \end{aligned} The equation for the tangent line that runs through (x_1, y_1) = \left(\dfrac{\pi}{3}, -\sqrt3\right) having a slope of m = 0 is \begin{aligned} y & = m(x-x_1)+y_1 \\ y & = 0\left(x-\dfrac{\pi}{6}\right) + (-\sqrt3) \\ y & = -\sqrt3 \end{aligned} [collapse] Problem Number 12 Evaluate the equation for the normal line on the curve of the following trigonometric functions at the given point. 1. h(\theta) = \theta + \sin \theta at the point whose ordinate is 0. 2. f(x) = x \cos x at the point whose abscissa is x = \dfrac{\pi}{3}. Solution Answer a) Given h(\theta) = \theta + \sin \theta. For y = 0, we have 0 = \theta + \sin \theta hence that \theta = 0. The tangent point is at (0, 0). The first derivative of function f(\theta)= \theta + \sin \theta is f'(\theta) = 1 + \cos \theta. The slope of the tangent line is represented as the value of function f’ at \theta = 0, that is m = f'(0) = 1 + \cos 0 = 2 The normal line is defined as the line that perpendicular to the tangent line at the point of tangency. The slope of the normal line is given by m_n = -\dfrac{1}{m} = -\dfrac12 The equation for the normal line that runs through (x_1, y_1) = (0, 0) having a slope of m_n = -\dfrac12 is \begin{aligned} y & = m_n(x-x_1)+y_1 \\ y & = -\dfrac12(x-0) + 0 \\ y & = -\dfrac12x \end{aligned} Thus, the equation for the normal line is given by \boxed{y = -\dfrac12x} Answer b) Given f(x) = x \cos x. For x = \dfrac{\pi}{3}, we have \begin{aligned} f\left(\dfrac{\pi}{3}\right) & = \dfrac{\pi}{3} \cos \dfrac{\pi}{3} \\ & = \dfrac{\pi}{3} \cdot \dfrac12 \\ & = \dfrac{\pi}{6} \end{aligned} The tangent point is at \left(\dfrac{\pi}{3}, \dfrac{\pi}{6}\right). The first derivative of function f(x) = x \cos x is f'(x) = \cos x-x \sin x (by applying the product rule). The slope of the tangent line is represented as the value of function f’ at x= \dfrac{\pi}{3}, that is \begin{aligned} m & = f’\left(\dfrac{\pi}{3}\right) \\ & = \cos \dfrac{\pi}{3}-\dfrac{\pi}{3} \sin \dfrac{\pi}{3} \\ & = \dfrac12-\dfrac{\pi}{3} \cdot \dfrac12\sqrt3 \\ & = \dfrac12-\dfrac{\sqrt3}{6}\pi \\ & = \dfrac{3-\sqrt3\pi}{6} \end{aligned} The normal line is defined as the line that perpendicular to the tangent line at the point of tangency. The slope of the normal line is given by m_n = -\dfrac{1}{m} = \dfrac{6}{\sqrt3\pi-3} The equation for the normal line that runs through (x_1, y_1) = \left(\dfrac{\pi}{3}, \dfrac{\pi}{6}\right) and bergradien m_n = \dfrac{6}{\sqrt3\pi-3} adalah \begin{aligned} y & = m_n(x-x_1)+y_1 \\ y & = \dfrac{6}{\sqrt3\pi-3}\left(x-\dfrac{\pi}{3}\right) + \dfrac{\pi}{6} \end{aligned} Thus, the equation for the normal line is given by \boxed{y = \dfrac{6}{\sqrt3\pi-3}\left(x-\dfrac{\pi}{3}\right) + \dfrac{\pi}{6}} [collapse] Problem Number 13 Find the first derivative of the following trigonometric functions. a. f(x) = \cos^7 (x^3) b. g(x) = \cot (2x^4) c. r(x) = (\sin 3x-9 \cos^3 x)^6 Solution Apply the chain rule several times. Answer a) Given f(x) = \cos^7 (x^3) = (\underbrace{\cos (x^3)}_{u})^7. We have \begin{aligned} f'(x) & = 7 (\cos^6 (x^3)) \cdot \underbrace{(-\sin (x^3)) \cdot 3x^2}_{u’} \\ & = -21x^2 \sin (x^3) \cos^6 (x^3) \end{aligned} Thus, the first derivative of function f(x) is \boxed{f'(x) = -21x^2 \sin (x^3) \cos^6 (x^3)} Answer b) Given g(x) = \cot (\underbrace{2x^4}_{u}). We have \begin{aligned} g'(x) & = -\csc^2 (2x^4) \cdot \underbrace{8x^3}_{u’} \\ & = -8x^3 \csc^2 (2x^4) \end{aligned} Thus, the first derivative of function g(x) is \boxed{g'(x) = -8x^3 \csc^2 (2x^4)} Answer c) Given r(x) = (\underbrace{\sin 3x-9 \cos^3 x}_{u})^6. We have r'(x) = 6 (\sin 3x-9 \cos^3 x)^5 \cdot \underbrace{(3 \cos 3x-27 \cos^2 x (-\sin x))}_{u’} Note: The derivative of v = \cos^3 x is v’ = 3 \cos^2 x (-\sin x). Thus, the first derivative of function r(x) is\boxed{r'(x) = 6 (\sin 3x-9 \cos^3 x)^5 \cdot (3 \cos 3x-27 \cos^2 x (-\sin x))}[collapse] Problem Number 14 Find the first derivative of the following functions. a. f(\theta) = \sin (\sin \theta) b. h(x) = \sin (2 \cos x) c. H(x) = \sin [\sin (\sin x)] Solution Apply the chain rule several times. Answer a) Given f(\theta) = \sin (\underbrace{\sin \theta}_{u}). We have \begin{aligned} f'(\theta) & = \cos (\sin \theta) \cdot \underbrace{\cos \theta}_{u’} \\ & = \cos \theta \cos (\sin \theta) \end{aligned} Answer b) Given h(x) = \sin (\underbrace{2 \cos x}_{u}). We have \begin{aligned} h'(x) & = \cos (2 \cos x) \cdot \underbrace{-2 \sin x}_{u’} \\ & = -2 \sin x \cos (2 \cos x) \end{aligned} Answer c) Given H(x) = \sin [\underbrace{\sin (\underbrace{\sin x}_{v})}_{u}]. We have \begin{aligned} H'(x) & = \cos [\sin (\sin x)] \cdot \underbrace{\cos (\sin x)}_{u’} \cdot \underbrace{\cos x}_{v’} \\ & = \cos x \cos (\sin x) \cos [\sin (\sin x)] \end{aligned} [collapse] Problem Number 15 By applying the chain rule, find the derivative of the following functions. a. y = \sin (4x^2+3x) b. y = (1+\sin^2 x)(1-\sin^2 x) c. y = \left(\dfrac{\cos x}{1+\sin x}\right)^3 d. y = \dfrac{(1+\sin x)^2}{x^3} e. y = (x^2-1)^3 \cos (3x+2) f. y = \dfrac{1}{\sqrt{\cos x^2}} Solution Answer a) Given y = \sin (\underbrace{4x^2+3x}_{u}). By applying the chain rule, we have its derivative. y’ = \underbrace{(8x + 3)}_{u’} \cos (4x^2+3x) Answer b) Given \begin{aligned} y & = (1+\sin^2 x)(1-\sin^2 x) \\ & = 1-\sin^4 x = 1-(\underbrace{\sin x}_{u})^4 \end{aligned} Its derivative is given by \begin{aligned} y’ & = 0-4 \sin^3 x \cdot \underbrace{\cos x}_{u} \\ & = -4 \sin^3 x \cos x \end{aligned} Answer c) Given y = \left(\underbrace{\dfrac{\cos x}{1+\sin x}}_{u}\right)^3. The derivative of u can be evaluated by applying the quotient rule, while for the function y, its derivative can be evaluated by applying the chain rule in whole.\begin{aligned} y’ & = 3 \cdot \dfrac{\cos x}{1+\sin x} \cdot \dfrac{(-\sin x)(1+\sin x)-(\cos x)(\cos x)}{(1+\sin x)^2} \\ & = \dfrac{3 \cos x(-\sin x-\sin^2 x-\cos^2 x)}{(1+\sin x)^3} \\ & = \dfrac{3 \cos x(-\sin x-1)}{(1+\sin x)^3} && (\sin^2 x + \cos^2 x = 1) \\ & = \dfrac{-3 \cos x\cancel{(1 + \sin x)}}{(1+\sin x)^{\cancelto{2}{3}}} \\ & = \dfrac{-3 \cos x}{(1+\sin x)^2} \end{aligned}Answer d) Given y = \dfrac{(1+\sin x)^2}{x^3}. Overall, we apply the quotient rule to find the derivative of y, but each of the components is differentiated by using the chain rule and product rule. Let: \begin{aligned} u & = (1+\sin x)^2 \\ \implies u’ & = 2(1+\sin x)(\cos x) \\ v & = x^3 \\ \implies v’ & = 3x^2 \end{aligned} Hence, its derivative is given by\begin{aligned} y’ & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{2(1+\sin x)(\cos x)(x^3)-(1+\sin x)^2(3x^2)}{(x^3)^2} \\ & = \dfrac{2x^3(1+\sin x)(\cos x)-3x^2(1+\sin x)^2}{x^6} \end{aligned}Answer e) Given y = (x^2-1)^3 \cos (3x+2). Overall, we apply the product rule to find the derivative of y, but each of the components is differentiated by using the chain rule. Let: \begin{aligned} u & = (x^2-1)^3 \\ \implies u’ & = 3(x^2-1)^2(2x) = 6x(x^2-1)^2 \\ v & = \cos (3x+2) \\ \implies v’ & = -3 \sin (3x+2) \end{aligned} Hence, its derivative is given by\begin{aligned} y’ & = u’v + uv’ \\ & = 6x(x^2-1)^2 \cos (3x+2) + (x^2-1)^3(-3 \sin (3x+2)) \\ & = (x^2-1)^2(6x \cos (3x+2)-3(x^2-1) \sin (3x+2)) \end{aligned}Answer f)
Given:
$y = \dfrac{1}{\sqrt{\cos x^2}} = (\underbrace{\cos x^2}_{u})^{-\frac12}$
By applying the power rule and the chain rule (to find the derivative of $u$), we have
\begin{aligned} y’ & = -\dfrac{1}{\cancel{2}}(\cos x^2)^{-\frac32} \cdot \underbrace{(-\sin x^2 \cdot \cancel{2}x)}_{u’} \\ & = \dfrac{x \sin x^2}{\cos x^2 \cdot \sqrt{\cos x}} \end{aligned}

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Problem Number 16 ($\bigstar$ HOTS $\bigstar$)
If $\dfrac{\text{d}f}{\text{d}x} = \dfrac{\sin x}{x}$ and $u(x) = \cot x$, evaluate $\dfrac{\text{d}f}{\text{d}u}$.

Solution

Given $\dfrac{\text{d}f}{\text{d}x} = \dfrac{\sin x}{x}$.
Since $u(x) = \cot x$, then we can express $x$ as the function of $u$ (recall the inverse function): $x = \text{arccot}~u$.
Note: Arcus (written as $\text{arc}$) is the notation to express the inverse of trigonometric functions.
The derivative of $x$ with respect to $u$ is
$\dfrac{\text{d}x}{\text{d}u} = -\dfrac{1}{1+x^2}$.

By applying the chain rule, we have
\begin{aligned} \dfrac{\text{d}f}{\text{d}u} & = \dfrac{\text{d}f}{\text{d}x} \cdot \dfrac{\text{d}x}{\text{d}u} \\ & = \dfrac{\sin x}{x} \cdot \left(-\dfrac{1}{1+x^2}\right) \\ & = -\dfrac{\sin x}{x + x^3} \end{aligned}

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