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

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can i go further?|dw:1336672014185:dw|
oh no you can't do that
no? :O why not????????????

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Other answers:

Because the exponent isn't constant
oh wait it is x^u where u is a funcrion?
Hi angela210793 :) Wrong attempt madam.
\[h=[f(x)]^{g(x)}\] \[\ln(h)=\ln([f(x)]^{g(x)})\] \[\ln(h)=g(x) \ln(f(x))\] Now differentiate both sides \[\frac{h'}{h}=g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\] \[h'=h(g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\]
ok..thanks Myini :D
of course assuming f>0
but whatever
@myininaya is this right? can i take this as a formula for every other problem similar to this one?[f%28x%29]^[g%28x%29]
@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^(g-1)f'+lnf*g'*f^g]

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