how do you find the derivative of (ln x)^sin x

2. Originally Posted by spazzyskylar
how do you find the derivative of (ln x)^sin x
one of two main approaches: try logarithmic differentiation. ever heard of it?

Let $y = (\ln x )^{\sin x}$

then $\ln y = \sin x \ln \ln x$

now differentiate both sides implicitly

3. Yeah, but every time I do it I screw it up..

4. Originally Posted by spazzyskylar
Yeah, but every time I do it I screw it up..
where are you messing up exactly? after my second line, what did you do?

5. For my answer I ended up with...

y'= y(sinx+cosx*ln(ln x)

6. Originally Posted by spazzyskylar
For my answer I ended up with...

y'= y(sinx+cosx*ln(ln x)
note that you need to differentiate ln(ln(x)) using the chain rule...

7. I thought I did in what I wrote down...

8. $y = (\ln{x})^{\sin{x}}$

$\ln{y} = \sin{x} \ln(\ln{x})$

$\frac{y'}{y} = \sin{x} \cdot \frac{\frac{1}{x}}{\ln{x}} + \ln(\ln{x}) \cos{x}$

$\frac{y'}{y} = \frac{\sin{x}}{x \ln{x}} + \ln(\ln{x}) \cos{x}$

$y' = (\ln{x})^{\sin{x}}\left[\frac{\sin{x}}{x \ln{x}} + \ln(\ln{x}) \cos{x}\right]$

9. Originally Posted by spazzyskylar
I thought I did in what I wrote down...
You have written down nothing. All you have writtten is an answer. Without the detailed working that goes with that answer we can only guess what mistakes you have made. Please show all your working, set out in an intelligible way, if you need more help.