Question

question 10 · 1 point
find the derivative of (h(x)= - 3cos^{13}(x)).
provide your answer below:
(h(x)=square)

Explanation:

Step1: Recall chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = \cos(x)$, so $h(x)=- 3u^{13}$.

Step2: Differentiate outer function

Differentiate $y=-3u^{13}$ with respect to $u$. Using the power - rule $\frac{d}{du}(au^{n})=nau^{n - 1}$, we get $\frac{dy}{du}=-3\times13u^{12}=-39u^{12}$.

Step3: Differentiate inner function

Differentiate $u = \cos(x)$ with respect to $x$. We know that $\frac{du}{dx}=-\sin(x)$.

Step4: Apply chain - rule

By the chain - rule $h'(x)=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = \cos(x)$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$. So $h'(x)=-39\cos^{12}(x)\times(-\sin(x)) = 39\sin(x)\cos^{12}(x)$.

Answer:

$39\sin(x)\cos^{12}(x)$