# Derivatives of exponential functions

Printable View

• Nov 29th 2009, 03:22 AM
ozgunatalay
Derivatives of exponential functions
• Nov 29th 2009, 03:33 AM
e^(i*pi)
Quote:

Originally Posted by ozgunatalay

This will require the chain rule and using logarithmic differentiation

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

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

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

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

Use the chain rule for each of them:

edit: oops no it isn't *edits*
edit2: that's better

Spoiler:

$\frac{1}{y} \, y' = \frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}$

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

$y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)$
• Nov 29th 2009, 04:35 AM
ozgunatalay
Thank you for your solution.
• Nov 29th 2009, 04:37 AM
e^(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, 04:48 AM
ozgunatalay
I am learning about this new. Sometimes when I'm having difficulty solving questions. Thanks for your suggestions.
• Nov 29th 2009, 06:02 AM
HallsofIvy
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, 06:25 AM
ozgunatalay
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, 10:35 AM
ozgunatalay
Is there anyone who can check if I did it right?
• Nov 29th 2009, 10:53 AM
hjortur
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:

$
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:

$
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, 11:51 AM
ozgunatalay
Where am I doing wrong?
• Nov 29th 2009, 11:54 AM
hjortur
Nowhere. Your answer is the same as the corrected one.
• Nov 29th 2009, 12:10 PM
ozgunatalay
• Nov 29th 2009, 12:41 PM
hjortur
Sorry (Happy), I didn't check out e^(i*pi)'s answer good enough.

But your answer is correct.
• Nov 29th 2009, 01:26 PM
ozgunatalay
Thank you for your interest in question (Happy)(Clapping)