# What is the derivative of inverse tangent of 2x?

Apr 14, 2015

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

Solution
Let

$y = {\tan}^{- 1} 2 x$

$\tan y = 2 x$

Differentiating both side with respect to 'x'

$\frac{d}{\mathrm{dx}} \left(\tan y\right) = \frac{d}{\mathrm{dx}} \left(2 x\right)$

$\implies {\sec}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2$

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

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{1 + {\tan}^{2} y}$

Now, as
$\tan y = 2 x$

${\tan}^{2} y = {\left(2 x\right)}^{2}$

${\tan}^{2} y = 4 {x}^{2}$

So,

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

Apr 14, 2015

dy/(dx)=2/(1+4x^2)#

Solution

Let

$y = {\tan}^{- 1} 2 x$

Differentiating both side with respect to 'x'

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {\left(2 x\right)}^{2}} .2 \frac{d}{\mathrm{dx}} \left(x\right)$

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

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

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