Tag: Aturan Rantai

Ingat kembali bahwa
$\begin{array}{rl}f(x)& =\sin x⟹f^{\prime }(x)=\cos x\\ f(x)& =\cos x⟹f^{\prime }(x)=-\sin x\end{array}$
Dengan menggunakan fakta di atas, kita peroleh
$\begin{array}{rl}y& =3\sin x-\cos x\\ ⟹y^{\prime }& =3\cos x-(-\sin x)\\ & =3\cos x+\sin x\end{array}$
Jadi, turunan dari $y=3\sin x-\cos x$ adalah $\boxed{{\displaystyle 3\cos x+\sin x}}$
(Jawaban B)

Ingat kembali bahwa
$f(x)=\cos x⟹f^{\prime }(x)=-\sin x$
Dengan menggunakan fakta di atas dan aturan turunan fungsi aljabar, kita peroleh
$$\begin{array}{rl}g(x)& =3x^2-{\displaystyle \frac{1}{2x^2}}+2\cos x\\ & =3x^2-{\displaystyle \frac{1}{2}}{x}^{-2}+2\cos x\\ g^{\prime }(x)& =3(2)x-{\displaystyle \frac{1}{2}}(-2)x^{-3}+2(-\sin x)\\ & =6x+{\displaystyle \frac{1}{x^3}}-2\sin x\end{array}$$Jadi, hasil dari $\boxed{{\displaystyle g^{\prime }(x)=6x+{\displaystyle \frac{1}{x^3}}-2\sin x}}$
(Jawaban A)

Ingat kembali bahwa
$\begin{array}{rl}f(x)& =\sin x⟹f^{\prime }(x)=\cos x\\ f(x)& =\cos x⟹f^{\prime }(x)=-\sin x\end{array}$
Dengan menggunakan fakta di atas, kita peroleh
$\begin{array}{rl}h(x)& =\sin x+\cos x\\ ⟹h^{\prime }(x)& =2\cos x+(-\sin x)\\ & =2\cos x-\sin x\end{array}$
Untuk $x={\displaystyle \frac{1}{2}}\pi$, kita peroleh
$\begin{array}{rl}{h}^{\prime }\left({\displaystyle \frac{1}{2}}\pi \right)& =2\cos {\displaystyle \frac{1}{2}}\pi -\sin {\displaystyle \frac{1}{2}}\pi \\ & =2(0)-1=-1\end{array}$
Jadi, nilai dari $\boxed{{\displaystyle h^{\prime }\left({\displaystyle \frac{1}{2}}\pi \right)=-1}}$
(Jawaban B)

Gunakan aturan hasil kali turunan.
$$\boxed{{\displaystyle f(x)=uv⟹f^{\prime }(x)=u^{\prime }v+u{v}^{\prime }}}$$Diketahui $T(x)=(\sin x+1)(\sin x-2).$
Misalkan:
$\begin{array}{rl}u& =\sin x+1⟹u^{\prime }=\cos x\\ v& =\sin x-2⟹v^{\prime }=\cos x\end{array}$
Dengan demikian, kita peroleh
$$\begin{array}{rl}{T}^{\prime }(x)& =u^{\prime }v+u{v}^{\prime }\\ & =(\cos x)(\sin x-2)+(\sin x+1)(\cos x)\\ & =\cos x\sin x-2\cos x+\cos x\sin x+\cos x\\ & =2\sin x\cos x-\cos x\\ & =\sin 2x-\cos x\end{array}$$Uretan: $\boxed{{\displaystyle \sin 2x=2\sin x\cos x}}$
Jadi, hasil diferensial (turunan) dari fungsi tersebut adalah $\boxed{{\displaystyle T^{\prime }(x)=\sin 2x-\cos x}}$
(Jawaban C)

Gunakan aturan hasil kali turunan.
$$\boxed{{\displaystyle f(x)=uv⟹f^{\prime }(x)=u^{\prime }v+u{v}^{\prime }}}$$Diketahui $h(\theta )=(\theta +{\displaystyle \frac{\pi }{2}})\sin \theta .$
Misalkan:
$\begin{array}{rl}u& =\theta +{\displaystyle \frac{\pi }{2}}⟹u^{\prime }=1\\ v& =\sin \theta ⟹v^{\prime }=\cos \theta \end{array}$
Dengan demikian, kita peroleh
$\begin{array}{rl}{h}^{\prime }(\theta )& =u^{\prime }v+u{v}^{\prime }\\ & =1(\sin \theta )+(\theta +{\displaystyle \frac{\pi }{2}})(\cos \theta )\\ & =\sin \theta +\theta \cos \theta +{\displaystyle \frac{\pi }{2}}\cos \theta \end{array}$
Jadi, hasil dari $\boxed{{\displaystyle h^{\prime }(\theta )=\sin \theta +\theta \cos \theta +{\displaystyle \frac{\pi }{2}}\cos \theta }}$
(Jawaban E)

Gunakan aturan hasil bagi turunan.
$$\boxed{{\displaystyle f(x)={\displaystyle \frac{u}{v}}⟹f^{\prime }(x)={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}}}$$Diketahui $g(\theta )={\displaystyle \frac{1-\sin \theta }{\sin \theta -3}}.$
Misalkan:
$\begin{array}{rl}u& =1-\sin \theta ⟹u^{\prime }=-\cos \theta \\ v& =\sin \theta -3⟹v^{\prime }=\cos \theta \end{array}$
Dengan demikian, kita peroleh
$$\begin{array}{rl}{g}^{\prime }(\theta )& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{(-\cos \theta )(\sin \theta -3)-(1-\sin \theta )(\cos \theta )}{(\sin \theta -3{)}^2}}\\ & ={\displaystyle \frac{\cancel{-\sin \theta \cos \theta }+3\cos \theta -\cos \theta +\cancel{\sin \theta \cos \theta }}{(\sin \theta -3{)}^2}}\\ & ={\displaystyle \frac{2\cos \theta }{(\sin \theta -3{)}^2}}\end{array}$$Jadi, hasil bagi diferensial (turunan) dari fungsi tersebut adalah $\boxed{{\displaystyle {\displaystyle \frac{2\cos \theta }{(\sin \theta -3{)}^2}}}}$
(Jawaban E)

Gunakan aturan hasil bagi turunan.
$$\boxed{{\displaystyle f(x)={\displaystyle \frac{u}{v}}⟹f^{\prime }(x)={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}}}$$Diketahui $R(t)={\displaystyle \frac{\sin t-\cos t}{\cos t+\sin t}}.$
Misalkan:
$$\begin{array}{rl}u& =\sin t-\cos t⟹u^{\prime }=\cos t+\sin t\\ v& =\cos t+\sin t⟹v^{\prime }=-\sin t+\cos t\end{array}$$Dengan demikian, kita peroleh
$$\begin{array}{rl}{R}^{\prime }(t)& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{(\cos t+\sin t)(\cos t+\sin t)-(\sin t-\cos t)(-\sin t+\cos t)}{(\cos t+\sin t{)}^2}}\\ & ={\displaystyle \frac{({\cos }^{2}t+\sin t\cos t+\sin t\cos t+{\sin }^{2}t)-(-{\sin }^{2}t+\sin t\cos t+\sin t\cos t-{\cos }^{2}t)}{{\cos }^{2}t+2\sin t\cos t+{\sin }^{2}t}}\\ & ={\displaystyle \frac{1+\cancel{2\sin t\cos t}-(-1)-\cancel{2\sin t\cos t}}{1+2\sin t\cos t}}\\ & ={\displaystyle \frac{2}{1+\sin 2t}}\end{array}$$Uretan: $\boxed{{\displaystyle \begin{array}{rl}\sin 2t& =2\sin t\cos t\\ {\sin }^{2}t+{\cos }^{2}t& =1\end{array}}}$
Jadi, turunan dari fungsi itu adalah $\boxed{{\displaystyle R^{\prime }(t)={\displaystyle \frac{2}{1+\sin 2t}}}}$
(Jawaban D)

Ingat kembali bahwa:
$\begin{array}{rl}f(x)& =\tan x⟹f^{\prime }(x)={\sec }^{2}x\\ f(x)& =\cot x⟹f^{\prime }(x)=-{\csc }^{2}x\end{array}$
Dengan menggunakan fakta di atas, kita peroleh
$\begin{array}{rl}y& =\tan x-\cot x\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\sec }^{2}x-(-{\csc }^{2}x)\\ & ={\sec }^{2}x+{\csc }^{2}x\end{array}$
Untuk $x={\displaystyle \frac{\pi }{4}},$ kita peroleh
$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}{|}_{x=\frac{\pi }{4}}& ={\sec }^2{\displaystyle \frac{\pi }{4}}+{\csc }^2{\displaystyle \frac{\pi }{4}}\\ & =(\sqrt{2}{)}^2+(\sqrt{2}{)}^2\\ & =2+2=4\end{array}$
Catatan: $\boxed{{\displaystyle \sec {\displaystyle \frac{\pi }{4}}=\csc {\displaystyle \frac{\pi }{4}}=\sqrt{2}}}$
Jadi, turunan pertama dari fungsi $y$ saat $x={\displaystyle \frac{\pi }{4}}$ adalah $\boxed{{\displaystyle 4}}$
(Jawaban E)

Ingat kembali bahwa:
$$\begin{array}{rl}f(x)& =\sec t⟹f^{\prime }(x)=\sec t\tan t\\ f(x)& =\csc t⟹f^{\prime }(x)=-\csc t\cot t\end{array}$$Dengan menggunakan fakta di atas, kita peroleh
$\begin{array}{rl}y& =\sec t-\csc t\\ y^{\prime }& =(\sec t\tan t)-(-\csc t\cot t)\\ & ={\displaystyle \frac{1}{\cos t}}\cdot {\displaystyle \frac{\sin t}{\cos t}}+{\displaystyle \frac{1}{\sin t}}\cdot {\displaystyle \frac{\cos t}{\sin t}}\\ & ={\displaystyle \frac{\sin t}{{\cos }^{2}t}}+{\displaystyle \frac{\cos t}{{\sin }^{2}t}}\\ & ={\displaystyle \frac{{\sin }^{3}t+{\cos }^{3}t}{(\sin t\cos t{)}^2}}\end{array}$
Jadi, turunan dari $y=\sec t-\csc t$ adalah $\boxed{{\displaystyle {\displaystyle \frac{{\sin }^{3}t+{\cos }^{3}t}{(\sin t\cos t{)}^2}}}}$
(Jawaban E)

Gunakan aturan hasil kali turunan.
$$\boxed{{\displaystyle f(x)=uv⟹f^{\prime }(x)=u^{\prime }v+u{v}^{\prime }}}$$Diketahui $y=x^3\tan x.$
Misalkan:
$\begin{array}{rl}u& =x^3⟹u^{\prime }=3x^2\\ v& =\tan x⟹v^{\prime }={\sec }^{2}x\end{array}$
Dengan demikian, kita peroleh
$\begin{array}{rl}{y}^{\prime }& =u^{\prime }v+u{v}^{\prime }\\ & =(3x^2)(\tan x)+(x^3)({\sec }^{2}x)\\ & =3x^2\tan x+(x^3)({\tan }^{2}x+1)\\ & =x^3{\tan }^{2}x+3x^2\tan x+x^3\end{array}$
Uretan: $\boxed{{\displaystyle {\sec }^{2}x={\tan }^{2}x+1}}$
Jadi, hasil dari $\boxed{{\displaystyle y^{\prime }=x^3{\tan }^{2}x+3x^2\tan x+x^3}}$
(Jawaban A)

Gunakan aturan hasil kali turunan.
$$\boxed{{\displaystyle f(x)=uv⟹f^{\prime }(x)=u^{\prime }v+u{v}^{\prime }}}$$Diketahui $h(x)=x^2\cot x.$
Misalkan:
$\begin{array}{rl}u& =x^2⟹u^{\prime }=2x\\ v& =\cot x⟹v^{\prime }=-{\csc }^{2}x\end{array}$
Dengan demikian, kita peroleh
$\begin{array}{rl}{h}^{\prime }(x)& =u^{\prime }v+u{v}^{\prime }\\ & =(2x)(\cot x)+(x^2)(-{\csc }^{2}x)\\ & =2x\cot x-x^2{\csc }^{2}x\end{array}$
Untuk $x={\displaystyle \frac{\pi }{4}}$, diperoleh
$$\begin{array}{rl}{h}^{\prime }\left({\displaystyle \frac{\pi }{4}}\right)& =2\cdot {\displaystyle \frac{\pi }{4}}\cdot \cot {\displaystyle \frac{\pi }{4}}-{\left({\displaystyle \frac{\pi }{4}}\right)}^2{\csc }^2{\displaystyle \frac{\pi }{4}}\\ & ={\displaystyle \frac{\pi }{2}}\cdot 1-{\displaystyle \frac{{\pi }^2}{16}}\cdot (\sqrt{2}{)}^2\\ & ={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{{\pi }^2}{8}}\\ & ={\displaystyle \frac{\pi }{8}}(4-\pi )\end{array}$$Jadi, nilai $\boxed{{\displaystyle h^{\prime }\left({\displaystyle \frac{\pi }{4}}\right)={\displaystyle \frac{\pi }{8}}(4-\pi )}}$
(Jawaban B)

Gunakan aturan hasil bagi turunan.
$$\boxed{{\displaystyle f(x)={\displaystyle \frac{u}{v}}⟹f^{\prime }(x)={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}}}$$Diketahui $g(x)={\displaystyle \frac{\cos x+2}{\sin x}}.$
Misalkan:
$\begin{array}{rl}u& =\cos x+2⟹u^{\prime }=-\sin x\\ v& =\sin x⟹v^{\prime }=\cos x\end{array}$
Dengan demikian, kita peroleh
$$\begin{array}{rl}{g}^{\prime }(x)& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{(-\sin x)(\sin x)-(\cos x+2)(\cos x)}{(\sin x{)}^2}}\\ & ={\displaystyle \frac{-{\sin }^{2}x-{\cos }^{2}x-2\cos x}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{-({\sin }^{2}x+{\cos }^{2}x)-2\cos x}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{-1-2\cos x}{{\sin }^{2}x}}\end{array}$$Untuk $x={\displaystyle \frac{\pi }{2}}$, diperoleh
$\begin{array}{rl}{g}^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)& ={\displaystyle \frac{-1-2\cos {\displaystyle \frac{\pi }{2}}}{{\sin }^2{\displaystyle \frac{\pi }{2}}}}\\ & ={\displaystyle \frac{-1-2(0)}{(1{)}^2}}\\ & ={\displaystyle \frac{-1-0}{1}}=-1\end{array}$
Jadi, nilai dari $\boxed{{\displaystyle g^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)=-1}}$
(Jawaban B)

Gunakan aturan hasil kali turunan.
$$\boxed{{\displaystyle f(x)=uv⟹f^{\prime }(x)=u^{\prime }v+u{v}^{\prime }}}$$Diketahui $f(x)=\sin x(2+\cos x).$
Misalkan:
$\begin{array}{rl}u=\sin x& ⟹u^{\prime }=\cos x\\ v=2+\cos x& ⟹v^{\prime }& =-\sin x\end{array}$
Dengan demikian, kita peroleh
$$\begin{array}{rl}{f}^{\prime }(x)& =u^{\prime }v+u{v}^{\prime }\\ & =(\cos x)(2+\cos x)+(\sin x)(-\sin x)\\ & =2\cos x+{\cos }^{2}x-{\sin }^{2}x\\ & =2\cos x+\cos 2x\end{array}$$Untuk $x={\displaystyle \frac{\pi }{4}}$, kita peroleh
$\begin{array}{rl}{f}^{\prime }\left({\displaystyle \frac{\pi }{4}}\right)& =2\cos {\displaystyle \frac{\pi }{4}}+\cos \cancel{2}\left({\displaystyle \frac{\pi }{{\cancel{4}}^2}}\right)\\ & =\cancel{2}\cdot {\displaystyle \frac{1}{\cancel{2}}}\sqrt{2}+0=\sqrt{2}\end{array}$
Jadi, nilai dari $\boxed{{\displaystyle f^{\prime }\left({\displaystyle \frac{\pi }{4}}\right)=\sqrt{2}}}$
(Jawaban C)

Diketahui $y=\cos (2x^3-x^4).$
Gunakan aturan rantai.
Misalkan $u=2x^3-x^4⟹u^{\prime }=6x^2-4x^3.$
Dengan demikian, didapat
$$\begin{array}{rl}y& =\cos u\\ ⟹y^{\prime }& =-\sin u\cdot u^{\prime }\\ & =-\sin (2x^3-x^4)\cdot (6x^2-4x^3)\\ & =-(6x^2-4x^3)\sin (2x^3-x^4)\end{array}$$Jadi, turunan pertama dari fungsi $y=\cos (2x^3-x^4)$ adalah $\boxed{{\displaystyle y^{\prime }=-(6x^2-4x^3)\sin (2x^3-x^4)}}$
(Jawaban E)

Diketahui $g(\theta )={\cos }^3\theta =(\underset{u}{\underbrace{\cos \theta }}{)}^3.$
Dengan menggunakan aturan rantai, diperoleh
$\begin{array}{rl}{g}^{\prime }(\theta )& =3{\cos }^2\theta \cdot \underset{u^{\prime }}{\underbrace{(-\sin \theta )}}\\ & =-3{\cos }^2\theta \sin \theta \end{array}$
Jadi, turunan fungsi tersebut adalah $\boxed{{\displaystyle g^{\prime }(\theta )=-3{\cos }^2\theta \sin \theta }}$
(Jawaban C)

Diketahui $y=\tan \underset{u}{\underbrace{(2\theta -3)}}.$
Dengan menggunakan aturan rantai, diperoleh
$\begin{array}{rl}{y}^{\prime }& ={\sec }^2(2\theta -3)\cdot \underset{u^{\prime }}{\underbrace{2}}\\ & =2{\sec }^2(2\theta -3)\end{array}$
Jadi, turunan fungsi tersebut adalah $\boxed{{\displaystyle y^{\prime }=2{\sec }^2(2\theta -3)}}$
(Jawaban D)

Diketahui $g(x)={\displaystyle \frac{\sin 2x-\cos x}{\cos 4x}}$.
Gunakan aturan hasil bagi.
Misalkan:
$\begin{array}{rl}u& =\sin 2x-\cos x\\ ⟹u^{\prime }& =2\cos 2x+\sin x\\ v& =\cos 4x\\ ⟹v^{\prime }& =-4\sin 4x\end{array}$
Dengan demikian, diperoleh
$$\begin{array}{rl}{g}^{\prime }(x)& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{(2\cos 2x+\sin x)(\cos 4x)-(\sin 2x-\cos x)(-4\sin 4x)}{(\cos 4x{)}^2}}\\ g^{\prime }\left({\displaystyle \frac{\pi }{4}}\right)& ={\displaystyle \frac{(2\cos {\displaystyle \frac{\pi }{2}}+\sin {\displaystyle \frac{\pi }{4}})(\cos \pi )-(\sin {\displaystyle \frac{\pi }{2}}-\cos {\displaystyle \frac{\pi }{4}})(-4\sin \pi )}{(\cos \pi {)}^2}}\\ & ={\displaystyle \frac{(2\cdot 0+{\displaystyle \frac{1}{2}}\sqrt{2})(-1)-(1-{\displaystyle \frac{1}{2}}\sqrt{2})(-4\cdot 0)}{(-1{)}^2}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{2}}\sqrt{2}-0}{1}}=-{\displaystyle \frac{1}{2}}\sqrt{2}\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle g^{\prime }\left({\displaystyle \frac{\pi }{4}}\right)=-{\displaystyle \frac{1}{2}}\sqrt{2}}}$
(Jawaban D)

Diketahui $h(\theta )={\sec }^4(p\theta +q)=(\underset{u}{\underbrace{\sec (p\theta +q)}}{)}^4.$
Dengan menggunakan aturan rantai, diperoleh
$$\begin{array}{rl}{h}^{\prime }(\theta )& =4{\sec }^3(p\theta +q)\cdot \underset{u^{\prime }}{\underbrace{\sec (p\theta +q)\tan (p\theta +q)\cdot p}}\\ & =4p\tan (p\theta +q){\sec }^4(p\theta +q)\\ & =4p\tan (p\theta +q)\cdot h(\theta )\end{array}$$Jadi, turunan pertama fungsi trigonometri tersebut adalah $\boxed{{\displaystyle h^{\prime }(\theta )=4p\tan (p\theta +q)\cdot h(\theta )}}$
(Jawaban B)

Diketahui $y=\sin 3x-\cos 3x$.
Dengan menggunakan aturan turunan dasar beserta aturan rantai, diperoleh
$\begin{array}{rl}{y}^{\prime }& =3\cos 3x-(-3\sin 3x)\\ & =3\cos 3x+3\sin 3x\end{array}$
Untuk $x={45}^{\circ }$, diperoleh
$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}{|}_{x={45}^{\circ }}& =3\cos 3({45}^{\circ })+3\sin 3({45}^{\circ })\\ & =3\cos {135}^{\circ }+3\sin {135}^{\circ }\\ & =3\cdot (-{\displaystyle \frac{1}{2}}\sqrt{2})+3\cdot {\displaystyle \frac{1}{2}}\sqrt{2}=0\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \frac{\text{d}y}{\text{d}x}}{|}_{x={45}^{\circ }}=0}}$
(Jawaban C)

Diketahui $f(x)=\underset{u}{\underbrace{x\sin x}}\underset{v}{\underbrace{\cos x}}$.
Gunakan aturan hasil kali dengan $u^{\prime }=1(\sin x)+x(\cos x)$ $=\sin x+x\cos x$ dan $v^{\prime }=-\sin x.$
Kita peroleh
$$\begin{array}{rl}{f}^{\prime }(x)& =u^{\prime }v+u{v}^{\prime }\\ & =(\sin x+x\cos x)(\cos x)+(x\sin x)(-\sin x)\\ & =\sin x\cos x+x{\cos }^{2}x-x{\sin }^{2}x\\ & ={\displaystyle \frac{1}{2}}(2\sin x\cos x)+x({\cos }^{2}x-{\sin }^{2}x)\\ & ={\displaystyle \frac{1}{2}}\sin 2x+x\cos 2x\end{array}$$Catatan: Ingat bahwa $\boxed{{\displaystyle \sin 2x=2\sin x\cos x}}$ dan $\boxed{{\displaystyle \cos 2x={\cos }^{2}x-{\sin }^{2}x}}$
Jadi, turunan dari $f(x)$ adalah $\boxed{{\displaystyle f^{\prime }(x)={\displaystyle \frac{1}{2}}\sin 2x+x\cos 2x}}$
(Jawaban A)

Diketahui $y=(\underset{a}{\underbrace{x\cos x}}{)}^2$.
Pertama, akan dicari turunan dari $a$ menggunakan aturan hasil kali.
Misalkan:
$\begin{array}{rl}u& =x⟹u^{\prime }=1\\ v& =\cos x⟹v^{\prime }=-\sin x\end{array}$
Untuk itu, diperoleh
$\begin{array}{rl}{a}^{\prime }& =u^{\prime }v+u{v}^{\prime }\\ & =1(\cos x)+x(-\sin x)\\ & =\cos x-x\sin x\end{array}$
Sekarang dengan aturan rantai, didapat
$\begin{array}{rl}{y}^{\prime }& =2a\cdot a^{\prime }\\ & =2(x\cos x)(\cos x-x\sin x)\\ & =2x{\cos }^{2}x-2x^2\sin x\cos x\end{array}$
Jadi, turunannya adalah $\boxed{{\displaystyle y^{\prime }=2x{\cos }^{2}x-2x^2\sin x\cos x}}$
(Jawaban A)