Author: Sukardi

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 $\boxed{3 \cos x+\sin x}$
(Answer B)

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, $\boxed{g'(x) = 6x+\dfrac{1}{x^3}-2 \sin x}$
(Answer A)

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 $\boxed{h’\left(\dfrac12\pi\right) = -1}$
(Answer B)

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 $\boxed{T'(x) =\sin 2x-\cos x}$
(Answer C)

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, $\boxed{h'(\theta) =\sin \theta + \theta \cos \theta + \dfrac{\pi}{2} \cos \theta}$
(Answer E)

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 $\boxed{\dfrac{2 \cos \theta}{(\sin \theta-3)^2}}$
(Answer E)

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 $\boxed{R'(t) = \dfrac{2}{1+ \sin 2t}}$
(Answer D)

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 $\boxed{4}$
(Answer E)

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 $\boxed{\dfrac{\sin^3 t + \cos^3 t}{(\sin t \cos t)^2}}$
(Answer E)

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, $\boxed{y’ = x^3 \tan^2 x + 3x^2 \tan x + x^3}$
(Answer A)

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 $\boxed{h’\left(\dfrac{\pi}{4}\right) = \dfrac{\pi}{8}(4-\pi)}$
(Answer B)

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 $\boxed{g’\left(\dfrac{\pi}{2}\right) = -1}$
(Answer B)

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 $\boxed{f’\left(\dfrac{\pi}{4}\right) = \sqrt2}$
(Answer C)

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 $\boxed{y’ = -(6x^2-4x^3) \sin (2x^3-x^4)}$
(Answer E)

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 $\boxed{g'(\theta) = -3 \cos^2 \theta \sin \theta}$
(Answer C)

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 $\boxed{y’= 2 \sec^2 (2\theta-3)}$
(Answer D)

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 $\boxed{g’\left(\dfrac{\pi}{4}\right) = -\dfrac12\sqrt2}$
(Answer D)

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 $\boxed{h'(\theta) = 4p \tan (p\theta+q) \cdot h(\theta)}$
(Answer B)

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 $\boxed{\dfrac{\text{d}y}{\text{d}x} \rvert_{x = 45^{\circ}} = 0}$
(Answer C)

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 $\boxed{f'(x) = \dfrac12 \sin 2x + x \cos 2x}$
(Answer A)

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 $\boxed{y’ = 2x \cos^2 x-2x^2 \sin x \cos x}$
(Answer A)

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 $\boxed{y’ = -\dfrac32(\cos (6x+9)-\cos (2x+3))}$
(Answer A)

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 $\boxed{y’ = \dfrac{\sin x \cos x}{\sqrt{1+\sin^2 x}}}$
(Answer E)

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
$y’_{x = 0} = \cos 0 \cdot \cos 0 \cdots \cos 0 = 1$
(Answer D)

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
$\boxed{f^{(2019})(x) = f^{\prime \prime \prime}(x) = -\cos x+\sin x}$
(Answer E)

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.

(Answer C)

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.

(Answer E)

Answer a)
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 $\boxed{\dfrac{\text{d}y}{\text{d}x} = 2 + \cos x}$
Answer b)
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 $\boxed{\dfrac{\text{d}y}{\text{d}x} = 5 \cos x + 6 \sin x}$
Answer c)
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 $\boxed{\dfrac{\text{d}y}{\text{d}x} = 24x^2-\cos x}$
Answer d)
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 $\boxed{\dfrac{\text{d}y}{\text{d}x} = -6 \sin x-16x-8}$
Answer e)
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 $\boxed{\dfrac{\text{d}y}{\text{d}x} = 2 \cos x + \sec^2 x}$
Answer f)
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 $\boxed{\dfrac{\text{d}y}{\text{d}x} = 2x-\sin x}$

Find the function’s derivative and then plug its variable as $\dfrac{\pi}{4}$.
Answer a)
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 $\boxed{f’\left(\dfrac{\pi}{4}\right) = 2\sqrt2+6}$
Answer b)
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 $\boxed{f’\left(\dfrac{\pi}{4}\right) = -\dfrac12\sqrt2}$
Answer c)
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, $\boxed{f’\left(\dfrac{\pi}{4}\right) = 0}$
Answer d)
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, $\boxed{f’\left(\dfrac{\pi}{4}\right) = 0}$
Answer e)
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, $\boxed{f’\left(\dfrac{\pi}{4}\right) = 4}$

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 $\boxed{x \cdot h'(x) = h(x) + x^2g'(x)}$

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 $\boxed{f'(x)= \dfrac{ad-bc}{(c \sin x + d \cos x)^2}}$

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
$\begin{aligned} k'(x) & = \dfrac{ad-bc}{(c \sin x + d \cos x)^2} \\ & = \dfrac{(-2)(5)-(4)(3)}{(3 \sin x + 5 \cos x)^2} \\ & = -\dfrac{22}{(3 \sin x + 5 \cos x)^2} \end{aligned}$

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)

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)

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 $\boxed{f'(t) = \dfrac{(6t^2-1) \sin^2 t + (6t^2-12) \cos^2 t}{(\sin t + 3t \cos t)^2}}$

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)

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 $\boxed{h'(y) = 3y^2 + y^2 \sin y}$

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
$\begin{aligned} y & = m(x-x_1)+y_1 \\ y & = 0\left(x-\dfrac{\pi}{6}\right) + (-\sqrt3) \\ y & = -\sqrt3 \end{aligned}$

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 $\boxed{y = \dfrac{6}{\sqrt3\pi-3}\left(x-\dfrac{\pi}{3}\right) + \dfrac{\pi}{6}}$ 

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 $$\boxed{r'(x) = 6 (\sin 3x-9 \cos^3 x)^5 \cdot (3 \cos 3x-27 \cos^2 x (-\sin x))}$$ 

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
$\begin{aligned} H'(x) & = \cos [\sin (\sin x)] \cdot \underbrace{\cos (\sin x)}_{u’} \cdot \underbrace{\cos x}_{v’} \\ & = \cos x \cos (\sin x) \cos [\sin (\sin x)] \end{aligned}$

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
$\begin{aligned} y’ & = -\dfrac{1}{\cancel{2}}(\cos x^2)^{-\frac32} \cdot \underbrace{(-\sin x^2 \cdot \cancel{2}x)}_{u’} \\ & = \dfrac{x \sin x^2}{\cos x^2 \cdot \sqrt{\cos x}} \end{aligned}$

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
$\begin{aligned} \dfrac{\text{d}f}{\text{d}u} & = \dfrac{\text{d}f}{\text{d}x} \cdot \dfrac{\text{d}x}{\text{d}u} \\ & = \dfrac{\sin x}{x} \cdot \left(-\dfrac{1}{1+x^2}\right) \\ & = -\dfrac{\sin x}{x + x^3} \end{aligned}$