Find $y^\prime$ if:
(a) $y=\tan ^{-1}( 4x)$ $\quad$ (b) $y=\sin ( \tan ^{-1}x)$

Solution (a) Let $y=\tan ^{-1}u$ and $u=4x.$ Then $\dfrac{dy}{du}=\dfrac{1}{1+u^{2}}$ and $\dfrac{du}{dx}=4.$ By the Chain Rule, $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\dfrac{1}{1+u^{2}} \cdot 4=\dfrac{4}{1+16x^{2}}$$

(b) Let $y=\sin u$ and $u=\tan ^{-1}x$. Then $\dfrac{dy}{du}=\cos u$ and $\dfrac{du}{dx}=\dfrac{1}{1+x^{2}}$. By the Chain Rule, $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\cos u\cdot \dfrac{1}{1+x^{2}}=\dfrac{\cos ( \tan ^{-1}x) }{1+x^{2}}$$