Question

How do you find the derivative of `lnsin^-1(x)`?

Answer 1

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

Explanation:

Use the derivative of natural logs rule:
`d/dx(lnu)=(u')/u`

So, find the derivative of `sin^-1x`, and divide it by `sin^-1x`.
The derivative of `sin^-1x` is `1/(sqrt(1-x^2))`
`d/dx(lnsin^-1x) = (1/(sqrt(1-x^2)))/(sin^-1x)`

After simplifying, we get
`d/dx(lnsin^-1x) = 1/((sin^-1x)sqrt(1-x^2))`

Answer 2

`\frac1{sin^-1(x)}\cdot\frac1\sqrt{1-x^2}`

Explanation:

This question is testing your knowledge of the chain rule.
The chain rule states that the derivative of a function`f(u)` where `u` is another function is`f'(u)\cdot\frac{du}dx`, so the derivative of `ln(sin^-1(x))` is `\frac1{sin^-1(x)}\cdot\fracddx[sin^-1(x)]`
And because `\fracddx[sin^-1(x)]=\frac1\sqrt{1-x^2}`,
`\fracddx[ln(sin^-1(x))]=\frac1{sin^-1(x)}\cdot\frac1\sqrt{1-x^2}`