• Jan 4th 2010, 04:39 PM
spazzyskylar
how do you find the derivative of (ln x)^sin x
• Jan 4th 2010, 04:41 PM
Jhevon
Quote:

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
• Jan 4th 2010, 04:42 PM
spazzyskylar
Yeah, but every time I do it I screw it up..
• Jan 4th 2010, 04:43 PM
Jhevon
Quote:

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?
• Jan 4th 2010, 04:53 PM
spazzyskylar
For my answer I ended up with...

y'= y(sinx+cosx*ln(ln x)
• Jan 4th 2010, 04:55 PM
Jhevon
Quote:

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...
• Jan 4th 2010, 05:22 PM
spazzyskylar
I thought I did in what I wrote down...
• Jan 4th 2010, 05:32 PM
skeeter
$y = (\ln{x})^{\sin{x}}$

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

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

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

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

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.