Question

Use logarithmic differentiation to find the derivative of the function.

y = (sin 3x)^lnx

Answer 1

Well, starting off, we just take the natural log of both sides so we can fix that power of ln(x)

\[\ln (y) = \ln(x)*\ln(\sin3x)\]Since I took the natural log, I was able to bring down that ln(x) power and make it into a multiplication instead. So now the right side is a product rule derivative. Would you know how to do that differentiation?

Answer 2

Tsk tsk, Psymon ofc.

Answer 3

Eh?

Answer 4

1/y = (ln(3sin(3x))/(x) +3ln(x)cot(3x) so how do I solve for y?

Answer 5

You sure you have your derivative correct?
\[f'(x)g(x) = f(x)g'(x)\]
f'(x) = (1/x)
g'(x) = 3cot(3x)
\[\frac{ 1 }{ y }=\frac{ \ln(\sin(3x)) }{ x }+3\cot(3x)\]

Answer 6

Oops, meant for that to be a plus in the formula, not an equal sign of course.

Answer 7

so now i have to solve for y right ?

Answer 8

Not really, no. When you do logarithmic differentiation, you're solving for dy/dx technically. What you do here is now multiply both sides by y to get:
\[\frac{ dy }{ dx }=y(\frac{ \ln(\sin3x) }{ x }+3\cot3x)\]Now we replace y with the original problem. When we started, we said y = (sin3x)^lnx. Well, thats exactly what we substitute y for. So that gives us:
\[\frac{ dy }{ dx }=(\sin3x)^{lnx}*(\frac{ \ln(\sin3x) }{ x }+3\cot3x)\]

And that would be your answer. There is no real simplifying that much.

Answer 9

Thanks for help .)

Answer 10

Yep. np :3