How do you take the derivative of tan^-1(x)?

Jul 30, 2018

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 x\right) = \frac{1}{1 + {x}^{2}} , x \in \mathbb{R}$

Explanation:

We know that ,

$\left({f}^{-} 1\right) ' \left(x\right) = \frac{1}{f ' \left({f}^{-} 1 \left(x\right)\right)} , f ' \left(x\right) \ne 0$

OR

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}}$

Let ,

${\tan}^{-} 1 x = a r c \tan x = y$ , where , $y \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

So , $x = \tan y$ , where , $x \in \mathbb{R}$

Diff. $x = \tan y$ w.r.t. $y$

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

Subst. $\tan y = x$

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

So,

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

$\therefore \frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 x\right) = \frac{1}{1 + {x}^{2}} , x \in \mathbb{R}$

Jul 30, 2018

The answer is $= \frac{1}{1 + {x}^{2}}$

Explanation:

Let

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

Then,

$\tan y = x$

Implicit differentiation yields

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y}$

But

$1 + {\tan}^{2} y = {\sec}^{2} y$

${\sec}^{2} y = 1 + {x}^{2}$

Finally,

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