Question

The function f(x) = sin^-1(tanx) is not differentiable at
(1)x=0
(2)X = -pie/6
(3) x = pie/6
(4)x = pie/4​

Answer 1

Answer:

1 is right answer

thank you ek bar check kar lena please answer ko

Answer 2

Given:

$f(x)=sin^{-1}(tanx)$

To find:

The value of x at which function is not differentiable.

Solution:

$f(x)=sin^{-1}(tanx)$

Differentiate w.r.t x

$f'(x)=\frac{1}{\sqrt{1-tan^2 x}}\times sec^2 x$

Using the formula

$\frac{d(tanx)}{dx}=sec^2 x$

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

$f'(x)=\frac{sec^2 x}{\sqrt{1-tan^2 x}}$

f'(x) is not differentiable when

$1-tan^2 x=0$

$tan^2x=1$

$tanx=\pm 1$

$tanx=1$

$tanx=tan\frac{\pi}{4}\implies x=\frac{\pi}{4}$

$tanx=-1$

$tanx=-tan\frac{\pi}{4}\implies tanx=tan(-\frac{\pi}{4})$

$x=-\frac{\pi}{4}$

By using $tan(-x)=-tanx$

Option (4) is true