## 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

1. angela210793 Group Title

can i go further?|dw:1336672014185:dw|

2. myininaya Group Title

oh no you can't do that

3. angela210793 Group Title

no? :O why not????????????

4. Ishaan94 Group Title

Because the exponent isn't constant

5. angela210793 Group Title

oh wait it is x^u where u is a funcrion?

6. angela210793 Group Title

function*

7. Shayaan_Mustafa Group Title

Hi angela210793 :) Wrong attempt madam.

8. myininaya Group Title

$h=[f(x)]^{g(x)}$ $\ln(h)=\ln([f(x)]^{g(x)})$ $\ln(h)=g(x) \ln(f(x))$ 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 Group Title

ok..thanks Myini :D

10. myininaya Group Title

of course assuming f>0

11. myininaya Group Title

but whatever

12. angela210793 Group Title

@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]

@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]