Question

if Y= Sin ^-1 √cosx Then Find dy/dx

Answer 1

we know,

if,

y = sin^-1 (f(x)

then,

dy/dx=[1/[ √1-(f(x))^2 ] ] .f'(X)

➡

so the given eqn is,

y=sin^-1 √cosx

Differentiatie with respect to x

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

$\frac{dy}{dx} = \frac{1}{ \sqrt{1 - ( { \sqrt{cosx} )}^{2} } } \times \frac{1}{2 \sqrt{cosx} } \\ \times ( - sinx)$

= -sinx/[2√cosx. √1-cosx]