A community for students.
Here's the question you clicked on:
 0 viewing
angela210793
 4 years ago
[(sinx)^cosx + (cosx)^sinx]'
Lemme write wht i think it may be the way to solve this....
angela210793
 4 years ago
[(sinx)^cosx + (cosx)^sinx]' Lemme write wht i think it may be the way to solve this....

This Question is Closed

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0can i go further?dw:1336672014185:dw

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.4oh no you can't do that

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0no? :O why not????????????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because the exponent isn't constant

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait it is x^u where u is a funcrion?

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0Hi angela210793 :) Wrong attempt madam.

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.4\[h=[f(x)]^{g(x)}\] \[\ln(h)=\ln([f(x)]^{g(x)})\] \[\ln(h)=g(x) \ln(f(x))\] Now differentiate both sides \[\frac{h'}{h}=g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\] \[h'=h(g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\]

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0ok..thanks Myini :D

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.4of course assuming f>0

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0@myininaya is this right? can i take this as a formula for every other problem similar to this one? http://www.wolframalpha.com/input/?i= [f%28x%29]^[g%28x%29]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@myininaya i think while differentiating f^g we can do this way, by assuming g as constant once and diff. f^g and then assume f as constant and diff f^g. now superimpose the two I always got correct answers with this method in fact we get [g*f^(g1)f'+lnf*g'*f^g]
Ask your own question
Sign UpFind more explanations on OpenStudy
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.