if $y = \\sin^{3}x$, then $\\frac{dy}{dx}=$\na $\\cos^{3}x$\nb $3\\cos^{2}x$\nc $3\\sin^{2}x$\nd…

if $y = \\sin^{3}x$, then $\\frac{dy}{dx}=$\na $\\cos^{3}x$\nb $3\\cos^{2}x$\nc $3\\sin^{2}x$\nd $3\\sin^{2}x\\cos x$\ne $3\\sin^{2}x\\cos x$

if $y = \\sin^{3}x$, then $\\frac{dy}{dx}=$\na $\\cos^{3}x$\nb $3\\cos^{2}x$\nc $3\\sin^{2}x$\nd $3\\sin^{2}x\\cos x$\ne $3\\sin^{2}x\\cos x$

Answer

Explanation:

Step1: Identify the outer - inner functions

Let $u = \sin x$, then $y = u^{3}$.

Step2: Differentiate outer function

The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=3u^{2}$.

Step3: Differentiate inner function

The derivative of $u=\sin x$ with respect to $x$ is $\frac{du}{dx}=\cos x$.

Step4: Apply chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = \sin x$, $\frac{dy}{du}=3u^{2}$ and $\frac{du}{dx}=\cos x$ into the chain - rule formula, we get $\frac{dy}{dx}=3\sin^{2}x\cos x$.

Answer:

D. $3\sin^{2}x\cos x$