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

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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:

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

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