let $f(x)=\tan^{-1}(cos(2x))$ $f(x)=$ question help: video message instructor submit question jump to answer

Explanation:

Step1: Apply chain - rule

Let $u = \cos(2x)$. Then $y=\tan^{- 1}(u)$. The derivative of $y = \tan^{-1}(u)$ with respect to $u$ is $\frac{dy}{du}=\frac{1}{1 + u^{2}}$, and the derivative of $u=\cos(2x)$ with respect to $x$ is $\frac{du}{dx}=-2\sin(2x)$.

Step2: Use the chain - rule formula $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u = \cos(2x)$ and the derivatives into the chain - rule formula. We get $f^{\prime}(x)=\frac{-2\sin(2x)}{1+\cos^{2}(2x)}$.

Answer:

$\frac{-2\sin(2x)}{1 + \cos^{2}(2x)}$