let ( f(x)=-8 cos left(cos left(x^{4}\right)\right) ). then ( f^{prime}(x)= )

Answer:

$32x^{3}\sin(\cos(x^{4}))\sin(x^{4})$

Explanation:

Step1: Apply the chain rule

Let $u = \cos(x^{4})$, then $f(x)=- 8\cos(u)$. The derivative of $y = -8\cos(u)$ with respect to $u$ is $y_u'=8\sin(u)$ (using the formula $(\cos t)'=-\sin t$).

Step2: Differentiate $u = \cos(x^{4})$

Let $v=x^{4}$, then $u = \cos(v)$. The derivative of $u$ with respect to $v$ is $u_v'=-\sin(v)$ (using the formula $(\cos t)'=-\sin t$), and the derivative of $v = x^{4}$ with respect to $x$ is $v_x' = 4x^{3}$ (using the power - rule $(x^n)'=nx^{n - 1}$).

Step3: Use the chain rule again

By the chain rule $f'(x)=y_u'\cdot u_v'\cdot v_x'$. Substitute $u = \cos(x^{4})$ and $v=x^{4}$: [ \begin{align*} f'(x)&=8\sin(\cos(x^{4}))\cdot(-\sin(x^{4}))\cdot4x^{3}\ &=32x^{3}\sin(\cos(x^{4}))\sin(x^{4}) \end{align*} ]