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

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

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

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
no? :O why not????????????
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Because the exponent isn't constant
 2 years ago

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

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
function*
 2 years ago

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

myininaya Group TitleBest 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 \[\frac{h'}{h}=g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\] \[h'=h(g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\]
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
ok..thanks Myini :D
 2 years ago

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

myininaya Group TitleBest ResponseYou've already chosen the best response.4
but whatever
 2 years ago

angela210793 Group TitleBest 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]
 2 years ago

vamgadu Group TitleBest 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]
 2 years ago
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