Question

Differentiate the function with respect to x.

`(sin x)^x + sin^(-1) sqrtx`

Differentiate the function `(sin x)^x + sin^(-1) sqrtx` with respect to x.

Answer 1

Let, y = `(sin x)^x + sin^-1 sqrtx`

Again, let y = u + v

Differentiating both sides with respect to x,

`dy/dx = (du)/dx + (dv)/dx`  ...(1)

∴ u = (sin x)^x

Taking logarithm of both sides,

log u = log (sin x)^x

log u = x log sin x

Differentiating both sides with respect to x,

`1/u (du)/dx = x d/dx log sin x + log sin x d/dx (x)`

= `x 1/(sin x) d/dx (sin x) + log sin x xx 1`

= `x * 1/(sin x) * cos x + log sin x xx 1`

= x cot x + log sin x

`therefore (du)/dx` = u (x cot x + log sin x)

= (sin x)^x [x cot x + log sin x]    ...(2)

Also, v = `sin^-1 sqrt x`

Differentiating both sides with respect to x,

`(dv)/dx = d/dx sin^-1 sqrtx`

= `1/sqrt(1 - x) d/dx x^(1/2)`

= `1/sqrt(1 - x) 1/2 x^(-1/2)`

= `1/(2sqrtx sqrt(1 - x))`  ...(3)

Putting the values ​​of `(du)/dx` and `(dv)/dx` from equations (2) and (3) in equation (1),

`(dy)/dx = (sin x)^x [x cot x + log sin x] +  1/(2sqrtx sqrt(1 - x))`