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- Nov 29th 2009, 02:22 AMozgunatalayDerivatives of exponential functions
- Nov 29th 2009, 02:33 AMe^(i*pi)
This will require the chain rule and using logarithmic differentiation

$\displaystyle y = \frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}{9}}}{ (x^3+6x)^{\frac{4}{11}}}$

$\displaystyle ln(y) = ln \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right)$

Since $\displaystyle ln \left(\frac{ab}{c}\right) = ln(a)+ln(b)-ln(c)$ we get

$\displaystyle ln(y) = ln \left[(x^2+2)^{\frac{3}{2}}\right] + ln \left[(x^2+9)^{\frac{4}{9}}\right] - ln \left[{(x^3+6x)^{\frac{4}{11}}}\right]$

Use the chain rule for each of them:

edit: oops no it isn't *edits*

edit2: that's better

__Spoiler__:

$\displaystyle y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)$ - Nov 29th 2009, 03:35 AMozgunatalay
Thank you for your solution.

- Nov 29th 2009, 03:37 AMe^(i*pi)
I suppose you could always use the quotient rule, the product rule and the chain rule but IMO that would lead to an even bigger mess

- Nov 29th 2009, 03:48 AMozgunatalay
I am learning about this new. Sometimes when I'm having difficulty solving questions. Thanks for your suggestions.

- Nov 29th 2009, 05:02 AMHallsofIvy
You might also want to note that there are NO exponential functions here. "Exponential functions" are functions that have the variable, x, in the exponent.

- Nov 29th 2009, 05:25 AMozgunatalay
I looked again and I ask questions, I reached a different result. If I've made a mistake, correct my mistakes.

http://img282.yukle.tc/images/2362turev22854.gif - Nov 29th 2009, 09:35 AMozgunatalay
Is there anyone who can check if I did it right?

- Nov 29th 2009, 09:53 AMhjortur
I can see no difference between your answer and e^(i*pi)'s, so I would conclude that this is correct.

EDIT: Ahh I see it now. e^(i*pi) did make a slight mistake in the last fraction:

$\displaystyle

y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)

$

is supposed to be:

$\displaystyle

y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x^2+24}{11(x^3+6x)^{\frac{4}{11}}}\right)

$ - Nov 29th 2009, 10:51 AMozgunatalay
Where am I doing wrong?

- Nov 29th 2009, 10:54 AMhjortur
Nowhere. Your answer is the same as the corrected one.

- Nov 29th 2009, 11:10 AMozgunatalay
- Nov 29th 2009, 11:41 AMhjortur
Sorry (Happy), I didn't check out e^(i*pi)'s answer good enough.

But your answer is correct. - Nov 29th 2009, 12:26 PMozgunatalay
Thank you for your interest in question (Happy)(Clapping)