Question

How do you differentiate `f(x) = sin(1/sqrt(arcsinx))` using the chain rule?

Answer 1

`-cos(1/sin^-1x)/(2sqrt(1-x^2)(sin^-1 x)^(3/2)`

Explanation:

let `u = sin^-1x`

`rArr sinu =x`

`rArr (dsinu)/dx = 1`

`cosu (du)/dx = 1`
`(du)/dx = 1/cosu = 1/(sqrt(1-sin^2u)`= `1/(sqrt(1-x^2)`

Now, take the derivative suing chain rule,
`(df(x))/dx= (df(u))/(du) (du)/(dx)`
`(dsin(1/sqrt(u)))/(dx)`=`d/(du)sin (1/sqrt(u)) (d u^(-½))/(du) (du)/(dx)`
`(df(x))/dx = cos (1/sqrt(u)) (-1/2u^(-3/2)) 1/(sqrt(1-x^2)`
`(df(x))/dx = -1/2 cos (1/sqrt(u))/(sqrt(1-x^2) u^(3/2)`
`(df(x))/dx=-cos(1/sin^-1x)/(2sqrt(1-x^2)(sin^-1 x)^(3/2)`