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

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angela210793Best ResponseYou've already chosen the best response.0
can i go further?dw:1336672014185:dw
 one year ago

myininayaBest ResponseYou've already chosen the best response.4
oh no you can't do that
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
no? :O why not????????????
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Because the exponent isn't constant
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
oh wait it is x^u where u is a funcrion?
 one year ago

Shayaan_MustafaBest ResponseYou've already chosen the best response.0
Hi angela210793 :) Wrong attempt madam.
 one year ago

myininayaBest 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 \[\frac{h'}{h}=g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\] \[h'=h(g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\]
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
ok..thanks Myini :D
 one year ago

myininayaBest ResponseYou've already chosen the best response.4
of course assuming f>0
 one year ago

angela210793Best 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]
 one year ago

vamgaduBest 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]
 one year ago
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