Question

question
find the derivative of $h(x)=2cos^{14}(x)$.
provide your answer below:
$h(x)=square$

Explanation:

Step1: Apply constant - multiple rule

The constant - multiple rule states that if $y = cf(x)$, then $y'=cf'(x)$. Here $c = 2$ and $f(x)=\cos^{14}(x)$. So $h'(x)=2\frac{d}{dx}(\cos^{14}(x))$.

Step2: Apply chain - rule

Let $u = \cos(x)$, then $y = u^{14}$. The chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=\frac{d}{du}(u^{14}) = 14u^{13}$ (using the power rule $\frac{d}{du}(u^n)=nu^{n - 1}$ with $n = 14$). Second, find $\frac{du}{dx}$: $\frac{du}{dx}=\frac{d}{dx}(\cos(x))=-\sin(x)$.

Step3: Substitute back and simplify

Since $u=\cos(x)$, we have $\frac{d}{dx}(\cos^{14}(x)) = 14\cos^{13}(x)\cdot(-\sin(x))=- 14\cos^{13}(x)\sin(x)$. Then $h'(x)=2\times(-14\cos^{13}(x)\sin(x))=-28\cos^{13}(x)\sin(x)$.

Answer:

$-28\cos^{13}(x)\sin(x)$