How do you find the derivative of #ln(arcsinx)^2#?

Question

2689 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-02T09:23:29

#2/(arcsinxsqrt(1-x^2))#

Given: #ln(arcsinx)^2#.

Use the chain rule, which states that,

#dy/dx=dy/(du)*(du)/dx#

Let #u=arcsinx,:.(du)/dx=1/(sqrt(1-x^2))#.

Then, #y=lnu^2=2lnu,:.dy/(du)=2/u#.

Combining, we get:

#dy/dx=2/u*1/(sqrt(1-x^2))#

#=2/(usqrt(1-x^2))#

Substitute back #u=arcsinx# to get the final answer:

#=2/(arcsinxsqrt(1-x^2))#