anonymous
  • anonymous
Use logarithmic differentiation to find the derivative of the function. y = (sin 3x)^lnx
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Psymon
  • Psymon
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?
Ness9630
  • Ness9630
Tsk tsk, Psymon ofc.
Psymon
  • Psymon
Eh?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
1/y = (ln(3sin(3x))/(x) +3ln(x)cot(3x) so how do I solve for y?
Psymon
  • Psymon
You sure you have your derivative correct? \[f'(x)g(x) = f(x)g'(x)\] f'(x) = (1/x) g'(x) = 3cot(3x) \[\frac{ 1 }{ y }=\frac{ \ln(\sin(3x)) }{ x }+3\cot(3x)\]
Psymon
  • Psymon
Oops, meant for that to be a plus in the formula, not an equal sign of course.
anonymous
  • anonymous
so now i have to solve for y right ?
Psymon
  • Psymon
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.
anonymous
  • anonymous
Thanks for help .)
Psymon
  • Psymon
Yep. np :3

Looking for something else?

Not the answer you are looking for? Search for more explanations.