find the derivative of the function. (g(\theta)=4cos^{4}(\theta)) (g(\theta)=)

Explanation:

Step1: Apply the constant - multiple rule

The constant - multiple rule states that if $y = cf(x)$, then $y'=cf'(x)$. Here $c = 4$ and $f(\theta)=\cos^{4}(\theta)$. So $g'(\theta)=4\frac{d}{d\theta}(\cos^{4}(\theta))$.

Step2: Apply the chain - rule

Let $u = \cos(\theta)$, then $y = u^{4}$. The chain - rule states that $\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=\frac{d}{du}(u^{4}) = 4u^{3}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$ with $n = 4$). Second, find $\frac{du}{d\theta}$: $\frac{du}{d\theta}=\frac{d}{d\theta}(\cos(\theta))=-\sin(\theta)$.

Step3: Substitute $u$ back and calculate $g'(\theta)$

Since $\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}$, and $u = \cos(\theta)$, we have $\frac{d}{d\theta}(\cos^{4}(\theta))=4\cos^{3}(\theta)\cdot(-\sin(\theta))$. Then $g'(\theta)=4\times4\cos^{3}(\theta)\cdot(-\sin(\theta))=-16\cos^{3}(\theta)\sin(\theta)$.

Answer:

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