Question

question find the derivative of $h(x)= - 3cos^{12}(x)$. provide your answer below: $h(x)=square$

Explanation:

Step1: Recall chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = \cos(x)$, so $H(x)=- 3u^{12}$.

Step2: Differentiate outer function

Differentiate $y=-3u^{12}$ with respect to $u$. Using the power rule $\frac{d}{du}(au^n)=nau^{n - 1}$, we get $\frac{d}{du}(-3u^{12})=-3\times12u^{11}=-36u^{11}$.

Step3: Differentiate inner function

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

Step4: Apply chain - rule

By the chain - rule $H^\prime(x)=\frac{dH}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dH}{du}=-36u^{11}$ and $\frac{du}{dx}=-\sin(x)$ back in, and replace $u$ with $\cos(x)$. So $H^\prime(x)=(-36\cos^{11}(x))\cdot(-\sin(x)) = 36\sin(x)\cos^{11}(x)$.

Answer:

$36\sin(x)\cos^{11}(x)$