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angela210793
 3 years ago
[(sinx)^cosx + (cosx)^sinx]'
Lemme write wht i think it may be the way to solve this....
angela210793
 3 years ago
[(sinx)^cosx + (cosx)^sinx]' Lemme write wht i think it may be the way to solve this....

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

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

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

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0Because the exponent isn't constant

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

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

myininaya
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok..thanks Myini :D

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

angela210793
 3 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]

vamgadu
 3 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]
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