angela210793
  • angela210793
[(sinx)^cosx + (cosx)^sinx]' Lemme write wht i think it may be the way to solve this....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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angela210793
  • angela210793
can i go further?|dw:1336672014185:dw|
myininaya
  • myininaya
oh no you can't do that
angela210793
  • angela210793
no? :O why not????????????

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anonymous
  • anonymous
Because the exponent isn't constant
angela210793
  • angela210793
oh wait it is x^u where u is a funcrion?
angela210793
  • angela210793
function*
Shayaan_Mustafa
  • Shayaan_Mustafa
Hi angela210793 :) Wrong attempt madam.
myininaya
  • myininaya
\[h=[f(x)]^{g(x)}\] \[\ln(h)=\ln([f(x)]^{g(x)})\] \[\ln(h)=g(x) \ln(f(x))\] 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
  • angela210793
ok..thanks Myini :D
myininaya
  • myininaya
of course assuming f>0
myininaya
  • myininaya
but whatever
angela210793
  • angela210793
@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
  • anonymous
@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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