Question

Find nth derivative of y tan inverse (1+x)/(1-x)

Answer 1

Answer:

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

Step-by-step explanation:

Given,

$\tan^{-1}\frac{(1 + x)}{1 - x}$

put,

$x = \tan\theta$

so,

$y = \tan^{-1}\frac{(1+\tan\theta)}{(1 - tan\theta)}$

$y = tan^-^1 (\frac {tan\frac{\pi}{4} + tan\theta}{1 - tan\frac{\pi}{4}.tan\theta})$

$y = \tan^{-1}({tan\frac{\pi}{4} + \theta })$

$y = \frac{\pi}{4} + \theta$

∵$x = \tan\theta$

⇒$\theta = tan^{-x}$

∴$y = \frac{\pi}{4} + tan^{-x}$

differentiate on both side,

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

its is also known as,

$\frac{dy}{dx} = \frac{1}{2i}\times(\frac{1}{x - i} - \frac{1}{x + i})$

       

Answer 2

Step-by-step explanation:

.........._____________

____________ __________________