You can solve this problem directly by differentiation rules but there’s an alternative simpler method of substitution!
Substitute
x = sin
in the given expression. Then

= sin
– 1x. As |x|

/2

/2.
dx = cos

d
d
/ dx = 1/cos
= sec
...(1)
Given expression becomes sin

.

/

+ log

=

sin

/ |cos

| + log|cos

|
=

sin

/ cos

+ logcos

[

for –

/2

/2, cos

> 0]
d(given expression) / dx = {d(given expression) / d

} {d

/ dx}
= {d(

tan

+ logcos

)/d

} {sec

} [From (1)]
= [

sec
2
+ tan

.1 + {1/cos

}{–sin

}] {sec

} [Apply uv rule of differentiation and formulae]
= [

sec
2
+ tan

– tan

] sec

=
sec3
...(2)
As x = sin

, x
2 = sin
2
= 1 – cos
2
.

cos
2
= 1 – x
2 i.e. sec
2
= 1/cos
2
= 1/(1 – x
2), sec

= 1/

Substitute the value of sec

in (1).
So, required answer is sin
– 1x . (1/

)
3 = sin
– 1x . 1/(1 – x
2)
3/2 = sin– 1x / (1 – x2)3/2 .