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angela210793

  • 2 years ago

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

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  1. angela210793
    • 2 years ago
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    can i go further?|dw:1336672014185:dw|

  2. myininaya
    • 2 years ago
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    oh no you can't do that

  3. angela210793
    • 2 years ago
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    no? :O why not????????????

  4. Ishaan94
    • 2 years ago
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    Because the exponent isn't constant

  5. angela210793
    • 2 years ago
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    oh wait it is x^u where u is a funcrion?

  6. angela210793
    • 2 years ago
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    function*

  7. Shayaan_Mustafa
    • 2 years ago
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    Hi angela210793 :) Wrong attempt madam.

  8. myininaya
    • 2 years ago
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    \[h=[f(x)]^{g(x)}\] \[\ln(h)=\ln([f(x)]^{g(x)})\] \[\ln(h)=g(x) \ln(f(x))\] Now differentiate both sides \[\frac{h'}{h}=g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\] \[h'=h(g'(x)\ln(f(x))+g(x)\frac{f'(x)}{f(x)})\]

  9. angela210793
    • 2 years ago
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    ok..thanks Myini :D

  10. myininaya
    • 2 years ago
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    of course assuming f>0

  11. myininaya
    • 2 years ago
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    but whatever

  12. angela210793
    • 2 years ago
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    @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]

  13. vamgadu
    • 2 years ago
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    @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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