# What is the derivative of tan^-1(4x)?

Apr 23, 2015

$\frac{4}{16 {x}^{2} + 1}$

Derivative of ${\tan}^{-} 1 4 x$ can be written using the formula for the derivative of ${\tan}^{-} 1 x$ = $\frac{1}{1 + {x}^{2}}$

$\frac{d}{\mathrm{dx}} {\tan}^{-} 1 4 x$ = $\frac{1}{16 {x}^{2} + 1} \frac{d}{\mathrm{dx}} \left(4 x\right)$

= $\frac{4}{16 {x}^{2} + 1}$

Apr 23, 2015

$y = \arctan \left(4 x\right)$

$\tan y = 4 x$

${\sec}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 4$

$\left({\tan}^{2} y + 1\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 4$

$\left(16 {x}^{2} + 1\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 4$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{16 {x}^{2} + 1}$