# Implicit Differentiation Question

• Sep 3rd 2009, 07:06 PM
xwrathbringerx
Implicit Differentiation Question
Hi

Could someone please show me how to differentiate y = [ln(x)]^(sinx)?

Thanxx a lot!
• Sep 3rd 2009, 07:17 PM
ynj
$(\ln x)^{\sin x}=e^{\ln \ln x\sin x}$
• Sep 4th 2009, 05:06 AM
Calculus26
y = ln(x)^(sin(x))

ln(y) = sin(x)ln(ln(x))

1/y dy/dx = cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x

dy/dx = y[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]

dy/dx = ln(x)^(sin(x))[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]
• Sep 4th 2009, 05:22 AM
xwrathbringerx
Quote:

Originally Posted by Calculus26
y = ln(x)^(sin(x))

ln(y) = sin(x)ln(ln(x))

1/y dy/dx = cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x

dy/dx = y[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]

dy/dx = ln(x)^(sin(x))[cos(x)ln(ln(x) + sin(x) (1/ln(x))1/x]

Can that part be simplified to:

sin(x) / (x ln(x))
= 1 / ln(x)
• Sep 4th 2009, 05:31 AM
Calculus26
How do you get sin(x)/[xln(x)] simplifying to 1/ln(x) ?

you are not suggesting sin(x)/x =1 are you?
• Sep 4th 2009, 05:34 AM
HallsofIvy
Quote:

Originally Posted by xwrathbringerx
Can that part be simplified to:

sin(x) / (x ln(x))
= 1 / ln(x)

$\lim_{x\rightarrow 0} \frac{sin(x)}{x}= 1$ but, in general, $\frac{sin(x)}{x}\ne 1$.