Question

differentiate of Sin inverse under root x​

Answer 1

Step-by-step explanation:

here, let us consider the function is

$y = { \sin }^{ - 1} \sqrt{x}$

Differentiation wrt x :

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

here remember that if the function is sin^(-1) f(x) then their derivative is( 1 / √[1 - (f (x))²]) x f'(x)

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

$= \frac{1}{ \sqrt{1 - x} } \times \frac{1}{2 \sqrt{x} }$

$= \frac{1}{2 \sqrt{x}( \sqrt{1 - x} ) }$

$\frac{dy}{dx} = \frac{1}{2 \sqrt{x}( \sqrt{1 - x} ) }$