if $f(x)=sin^{3}x$, find $f(x)$\nfind $f(3)$

Explanation:

Step1: Apply chain - rule

Let $u = \sin x$, then $y = u^{3}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. $\frac{dy}{du}=\frac{d(u^{3})}{du}=3u^{2}$ and $\frac{du}{dx}=\frac{d(\sin x)}{dx}=\cos x$.

Step2: Substitute $u$ back

Substitute $u = \sin x$ into $\frac{dy}{du}\cdot\frac{du}{dx}$. So $f^{\prime}(x)=3\sin^{2}x\cdot\cos x$.

Step3: Find $f^{\prime}(3)$

Substitute $x = 3$ into $f^{\prime}(x)$. So $f^{\prime}(3)=3\sin^{2}(3)\cos(3)$.

Answer:

$f^{\prime}(x)=3\sin^{2}x\cos x$ $f^{\prime}(3)=3\sin^{2}(3)\cos(3)$