Find \(y^\prime \) if:

\(y=\tan ^{-1}( 4x)\)

\(y=\sin ( \tan ^{-1}x)\)

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, \[ y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\cos u\cdot \dfrac{1}{1+x^{2}}=\dfrac{\cos ( \tan ^{-1}x) }{1+x^{2}} \]