Is This the Correct Method for Finding dy/dx of y = ln(sin^{-1}(x))?

[monet A]

To find dy/dx for

$$y = ln(sin^{-1}(x))$$

I did this:

$$y = ln(sin^{-1}(x))$$

so

$$e^y = (sin^{-1}(x))$$

and

$$e^y \frac {dy}{dx} = \frac {1} {\sqrt{1-x^2}}$$

then

$$\frac {dy}{dx} = \frac {1}{\sqrt {1-x^2} * e^y}$$

$$= \frac {1}{\sqrt {1-x^2} * e^{ln(sin^{-1}(x))}}$$

$$= \frac {1}{\sqrt {1-x^2} * (sin^{-1}(x))}}$$

I think that its all valid implicit differentiation but I m not 100% confident about it, please help.

 

[hypermorphism]

It looks fine to me, and the result checks out with ordinary differentiation. Which part were you uncertain about ?

 

[Zurtex]

You could have just as easily applied the chain rule but that seems to be fine

 

[monet A]

It looks fine to me, and the result checks out with ordinary differentiation. Which part were you uncertain about ?

Just the validity of transferring y to differentiate both sides, I wasn't sure that I was differentiating the same equation, probably over thinking it.

Thanks heaps though because I think I am starting to become more capable with it but I don't want to be overconfident in mistakes.