Finding dy/dx for y = ln(sin^{-1}(x))

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monet A
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To find dy/dx for

[tex]y = ln(sin^{-1}(x))[/tex]

I did this:

[tex] y = ln(sin^{-1}(x)) [/tex]

so

[tex] e^y = (sin^{-1}(x)) [/tex]

and

[tex] e^y \frac {dy}{dx} = \frac {1} {\sqrt{1-x^2}} [/tex]

then

[tex] \frac {dy}{dx} = \frac {1}{\sqrt {1-x^2} * e^y} [/tex]

[tex] = \frac {1}{\sqrt {1-x^2} * e^{ln(sin^{-1}(x))}} [/tex]

[tex] = \frac {1}{\sqrt {1-x^2} * (sin^{-1}(x))}} [/tex]

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

:frown:
 
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It looks fine to me, and the result checks out with ordinary differentiation. Which part were you uncertain about ?
 
You could have just as easily applied the chain rule but that seems to be fine :smile:
 
hypermorphism said:
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.

:cool: