Have I done this right?

  • Thread starter 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:
 
Last edited:
It looks fine to me, and the result checks out with ordinary differentiation. Which part were you uncertain about ?
 

Zurtex

Science Advisor
Homework Helper
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You could have just as easily applied the chain rule but that seems to be fine :smile:
 
67
0
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:
 

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