Question

Differentiate the function w.r.t.x. : $\sin^{-1} ( \frac{1-25x^{2}} {1+25x^{2}} )$

Answer 1

Answer:

$\frac {d}{dx}sin^{-1} (\frac{1-25x^{2} }{1+25x^{2} } )=\frac{-10}{1+25x^{2} }$

Step-by-step explanation:

To solve this without much difficulty we must use substitution method of integration

let 5x = tan θ

$\theta= tan^{-1} 5x$

let

$y=sin^{-1} (\frac{1-25x^{2} }{1+25x^{2} } )$

substitute the value of 5x

$y=sin^{-1} (\frac{1-tan^{2}\theta }{1+tan^{2}\theta } )\\ \\ =sin^{2}(cos 2\theta)\\ \\ \\ since\\ \\ cos 2\theta= \frac{1-tan^{2}\theta }{1+tan^{2}\theta } \\\\ \\ and\:\: we \:\:\:\:know\:\:\: that\\ \\ cos\theta = sin(90-\theta)$

$= sin^{2}[sin( \frac{\pi}{2} - 2\theta)]$

since from yhe properties of inverse trigonometric functions

$y=\frac{\pi}{2} -2\theta$

put the substituted value

$y=\frac{\pi}{2} -2tan^{-1} 5x\\ \\ \frac{dy}{dx} =0-2\frac{1}{1+(5x)^{2} } .5\\ \\ \\ =\frac{-10}{1+25x^{2} }$

Answer 2

Answer:  $\frac{-10}{1+25x^2}$

Step-by-step explanation:

Let,

$y=\sin^{-1}(\frac{1-25x^2}{1+25x^2})\\\;\\Putting\;5x=\tan\theta\implies25x^2=\tan^2\theta\\\;\\y=\sin^{-1}(\frac{1-\tan^2\theta}{1+\tan^2\theta})\\\;\\y=\sin^{-1}(\cos2\theta)\\\;\\y=\sin^{-1}(\sin(\frac{\pi}{2}-2\theta))\\\;\\y=\frac{\pi}{2}-2\theta\\\;\\y=\frac{\pi}{2}-2\tan^{-1}(5x)\\\;\\\text{Diff. both sides w.r.t. x}\\\;\\\frac{dy}{dx}=0-2\frac{1}{1+(5x)^2}.\frac{d(5x)}{dx}\\\;\\\frac{dy}{dx}=\frac{-10}{1+25x^2}$

Note:-

$1.\cos2\theta=\frac{1-tan^2\theta}{1+\tan^2\theta}\\\;\\2.\frac{d(\tan x)}{dx}=\frac{1}{1+x^2}$