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

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- Jan 4th 2010, 04:39 PMspazzyskylarDerivative help, please
how do you find the derivative of (ln x)^sin x

- Jan 4th 2010, 04:41 PMJhevon
- Jan 4th 2010, 04:42 PMspazzyskylar
Yeah, but every time I do it I screw it up..

- Jan 4th 2010, 04:43 PMJhevon
- Jan 4th 2010, 04:53 PMspazzyskylar
For my answer I ended up with...

y'= y(sinx+cosx*ln(ln x) - Jan 4th 2010, 04:55 PMJhevon
- Jan 4th 2010, 05:22 PMspazzyskylar
I thought I did in what I wrote down...

- Jan 4th 2010, 05:32 PMskeeter
$\displaystyle y = (\ln{x})^{\sin{x}}$

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

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

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

$\displaystyle y' = (\ln{x})^{\sin{x}}\left[\frac{\sin{x}}{x \ln{x}} + \ln(\ln{x}) \cos{x}\right]$ - Jan 4th 2010, 05:34 PMmr fantastic