# Differentiate it please as earlier as possible and explain it briefly

Samyak Jain
333 Points
5 years ago
You can solve this problem directly by differentiation rules but there’s an alternative simpler method of substitution!
Substitute x = sin$\theta$ in the given expression. Then $\theta$ = sin– 1x. As |x| $\pi$/2 $\theta$ $\pi$/2.
dx = cos$\theta$ d$\theta$  $\dpi{80} \Rightarrow$  d$\theta$ / dx = 1/cos$\theta$ = sec$\theta$          ...(1)
Given expression becomes sin$\theta$.$\theta$ / $\dpi{80} \sqrt{1-sin^2\theta}$  +  log$\dpi{80} \sqrt{1-sin^2\theta}$   =   $\theta$ sin$\theta$ / |cos$\theta$|  +  log|cos$\theta$|
=  $\theta$ sin$\theta$ / cos$\theta$  +  logcos$\theta$     [$\dpi{80} \because$ for – $\pi$/2 $\theta$ $\pi$/2,  cos$\theta$ > 0]
= $\theta$ tan$\theta$ + logcos$\theta$
d(given expression) / dx  =  {d(given expression) / d$\theta$} {d$\theta$ / dx}
= {d($\theta$ tan$\theta$ + logcos$\theta$)/d$\theta$} {sec$\theta$}             [From (1)]
= [$\theta$ sec2$\theta$ + tan$\theta$ .1 + {1/cos$\theta$}{–sin$\theta$}] {sec$\theta$}     [Apply uv rule of differentiation and formulae]
= [$\theta$ sec2$\theta$ + tan$\theta$ – tan$\theta$] sec$\theta$
=  $\theta$ sec3$\theta$                ...(2)
As x = sin$\theta$,  x2 = sin2$\theta$ = 1 – cos2$\theta$.
$\dpi{80} \therefore$ cos2$\theta$ = 1 – x2   i.e.   sec2$\theta$ = 1/cos2$\theta$ = 1/(1 – x2), sec$\theta$ = 1/$\dpi{80} \sqrt{1-x^2}$
Substitute the value of sec$\theta$ in (1).
So, required answer is sin– 1x . (1/$\dpi{80} \sqrt{1-x^2}$)3  =  sin– 1x . 1/(1 – x2)3/2
= sin– 1x / (1 – x2)3/2 .