Question

Find
dy/dx
if y=cos^-1
$\sqrt{cos \: x}$

​

Answer 1

Question :

Find dy/dx if

y = cos⁻¹(√cosx)

Theory and Formulas of Differention:

•Chain rule

Let y =f(t) , t = g(u) and u =m(x)

Then ,

$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{du} \times \frac{du}{dx}$

•some genral derivatives

$1) \frac{d( \cos(x) )}{dx} = - sin \: x$

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

$3) \frac{dy( \cos {}^{ - 1} (x)) }{dx} = \frac{ - 1}{ \sqrt{1 - x {}^{2} } }$

$4) \frac{d( \cot {}^{ - 1} (x)) }{dx} = \frac{ - 1}{1 + x {}^{2} }$

$5) \frac{d( \tan {}^{ - 1} (x) )}{dx} = \frac{1}{1 + x {}^{2} }$

$6) \frac{d( \sqrt{x} )}{dx} = \frac{1}{2 \sqrt{x} }$

Solution :

$y = \cos {}^{ - 1} ( \sqrt{ \cos \: x} )$

use chain rule , to Differnatiate the term

$\frac{dy}{dx} = \frac{d( \cos {}^{ - 1} \sqrt{ cos \: x} ) }{d( \sqrt{ \cos \: x} )} \times \frac{d( \sqrt{ \cos \: x } )}{d(cos \: x)} \times \frac{d(cosx)}{dx}$

$\implies \: \frac{dy}{dx} = \frac{ - 1}{ \sqrt{1 - \cos(x) } } \times \frac{1}{2 \sqrt{cos \: x} } \times - sin \: x$

$\frac{dy}{dx} = \frac{sin \: x}{2 \sqrt{cos \: x} \times \sqrt{1 - \cos(x) } }$

which is the required solution!

Answer 2

$\large \underline{ \red{ \boxed{ \bf \orange{Required \: Answer}}}}$