Question

question
find the derivative of $h(x)=-sin(2x^{4}-3)$.
provide your answer below:
$h(x)=square$

Explanation:

Step1: Apply chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = 2x^{4}-3$, so $h(x)=-\sin(u)$.

Step2: Find derivative of outer function

The derivative of $y =-\sin(u)$ with respect to $u$ is $y^\prime_{u}=-\cos(u)$.

Step3: Find derivative of inner function

The derivative of $u = 2x^{4}-3$ with respect to $x$ is $u^\prime_{x}=8x^{3}$.

Step4: Apply chain - rule formula

By the chain - rule $h^\prime(x)=y^\prime_{u}\cdot u^\prime_{x}$. Substitute $y^\prime_{u}=-\cos(u)$ and $u^\prime_{x}=8x^{3}$ back in, and replace $u$ with $2x^{4}-3$. So $h^\prime(x)=-\cos(2x^{4}-3)\cdot8x^{3}=- 8x^{3}\cos(2x^{4}-3)$.

Answer:

$-8x^{3}\cos(2x^{4}-3)$