Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Logarithmic differantiation,
http://screencast.com/t/fwBW2D9z9E
Can you help me out on this please
 10 months ago
 10 months ago
Logarithmic differantiation, http://screencast.com/t/fwBW2D9z9E Can you help me out on this please
 10 months ago
 10 months ago

This Question is Closed

reemiiBest ResponseYou've already chosen the best response.0
when there is an \(x\) in the exponent and want to compute a limit or a derivative, always resort to this technique: \[ a^{h(x)} = e^{\ln a^{h(x)}} = e^{h(x) \ln a} \] from here you should be able to do something.
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
were you able to do it??
 10 months ago

reemiiBest ResponseYou've already chosen the best response.0
do you know the "more simple" formula: \((a^x)' = a^x \ln a\) ? (\(a>0\)) knowing this you don't need the above trick. (you have to know the derivation of a compound function formula)
 10 months ago

.Sam.Best ResponseYou've already chosen the best response.1
Let y=f(x) \[\Large y=\pi^{sinx}\] ln both sides \[\ln(y)=\sin(x)\ln(\pi)\] Differentiate each term \[\frac{1}{y}y'=\ln(\pi)\cos(x) \\ \\ y'=yln(\pi)\cos(x) \\ \\ y'=\pi^{sinx}\ln(\pi)\cos(x)\] Got it?
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
If that's it yea easy enough
 10 months ago

reemiiBest ResponseYou've already chosen the best response.0
("always" means that it works, but there can be simpler ways depending on the problem) \[ (\pi^{\sin x})' = (\pi^{\sin x} \ln \pi) (\sin x)' = (\pi^{\sin x} \ln \pi) (\cos x) \] (derivation by parts.)
 10 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.