Question

find the derivative of the function.
g(\theta)=4\cos^{4}(\theta)
g(\theta)=16\sin^{3}\cdot| your answer cannot be

Explanation:

Step1: Apply chain - rule

Let $u = \cos(\theta)$, then $g(\theta)=4u^{4}$. The derivative of $y = 4u^{4}$ with respect to $u$ is $\frac{dy}{du}=16u^{3}$ by the power - rule $\frac{d}{du}(au^{n})=nau^{n - 1}$ where $a = 4$ and $n = 4$. The derivative of $u=\cos(\theta)$ with respect to $\theta$ is $\frac{du}{d\theta}=-\sin(\theta)$.

Step2: Use chain - rule formula

By the chain - rule $\frac{dg}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}$. Substitute $\frac{dy}{du}=16u^{3}$ and $\frac{du}{d\theta}=-\sin(\theta)$ back in, and replace $u$ with $\cos(\theta)$. So $\frac{dg}{d\theta}=16\cos^{3}(\theta)\cdot(-\sin(\theta))=- 16\sin(\theta)\cos^{3}(\theta)$.

Answer:

$-16\sin(\theta)\cos^{3}(\theta)$