# Have I done this right?

1. Jun 22, 2005

### 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

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

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

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

Last edited: Jun 22, 2005
2. Jun 22, 2005

### hypermorphism

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

3. Jun 22, 2005

### Zurtex

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

4. Jun 22, 2005

### monet A

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.