Question

Sin^-1(lnx) find the derivative

Answer 1

Use chain rule to split the problem into two:

`(d)/(dx)\sin^(-1)(\ln x)=(d sin^(-1)u)/(du)(du)/(dx)`

with
`u = \ln x`

Let

`y = \sin^(-1)u`

since sin^-1 is an inverse to sin we also know that

`\sin y = u`

We are looking for

`(dy)/(du) = (d \sin^(-1)u)/(du)\\(1)/((du)/(dy))= (d \sin^(-1)u)/(du)\\(1)/(\cos y)= (d \sin^(-1)u)/(du)\\(1)/(√(1-sin^2 y))= (d \sin^(-1)u)/(du)\\(1)/(√(1-u^2))= (d \sin^(-1)u)/(du)`

Now the second part of the chain rule:

`(du)/(dx)=(\ln x)/(dx)=(1)/(|x|)=(1)/(x)\,\,\mbox{for}\,\,x>0`

and putting it all together

`(d sin^(-1)u)/(du)(du)/(dx)= (1)/(x√(1-\ln^2 x))`

the last form is the final derivative and your answer