Question

Find the differential coefficient of cos-¹x² with respect to x​

Answer 1

Question:-

Find the differential coefficient of with respect to x.

Given:-

• $\sf \large cos-¹x²$

To Find:-

• Differentiation of w.r.t.x.

Solution:-

$\sf \large Let, y = cos-¹x²$

$\sf \large \therefore \frac{dy}{dx} = \frac{d}{dx} (cos-¹x²)$

$\sf \large = - \frac{1}{ \sqrt{1 - (x {}^{2} ) { }^{2} } } . \frac{d}{dx} (x {}^{2} ) \: \: ( \frac{d}{dx} .cos {}^{ - 1} (x) = \frac{ - 1}{ \sqrt{1 - x {}^{2} } } )$

$\sf \large = \frac{ - 1}{ \sqrt{1 - x {}^{4} } } .2x$

$\sf \large = \frac{ - 2x}{ \sqrt{1 - x {}^{4} } }$

Answer:-

$\sf { \boxed{ \sf \purple{Hence, \: Differentiation \: of \: cos-¹x² \: = \frac{ - 2x}{ \sqrt{1 - x {}^{4} } }.}}}$

Hope you have satisfied. ⚘

Answer 2

Given,

function is $cos^{-1} x^{2}$

To Find,

the differential coefficient of $cos^{-1} x^{2}$  w.r.t 'x'.

Solution,

given function is $cos^{-1} x^{2}$

consider that y = $cos^{-1} x^{2}$                              (1)

now, differentiating both sides of equation (1) w.r.t 'x'

$\frac{d}{dx}y= \frac{d}{dx} cos^{-1} x^{2}$

we know that,

$\frac{d}{dx} cos^{-1} x^{2}=-\frac{1}{\sqrt{1-x^{2} } }$

and

using  $[\frac{d}{dx}f(g(x))=f'[g(x)]\frac{d}{dx}g(x)]$

we get,

$\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{4} } }\frac{d}{dx}x^{4}$

also, we know,

$\frac{d}{dx}x^{n} =nx^{n-1}$

so using the above result we get,

$\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{4} } }(2x)$

$\frac{d}{dx}cos^{-1} x^{2} =\frac{-2x}{\sqrt{1-x^{4} } }$

Hence the differential coefficient of $cos^{-1} x^{2}$ with respect to x is $\frac{-2x}{\sqrt{1-x^{4} } }$ .