Answer
$\dfrac {4x}{1+16x^{2}}+\tan ^{-1}4x$
Work Step by Step
$\dfrac {d}{dx}\left( x\tan ^{-1}\left( 4x\right) \right) =\left( \dfrac {d}{dx}\left( x\right) \right) x\tan ^{-1}\left( 4x\right) +xx\left( \dfrac {d}{dx}\tan ^{-1}\left( 4x\right) \right) =\tan ^{-1}4x+\dfrac {x}{1+\left( 4x\right) ^{2}}\times \dfrac {d}{dx}\left( 4x\right) =\dfrac {4x}{1+16x^{2}}+\tan ^{-1}4x$
