Derivatives of exponential functions

**e^(i*pi)** · November 29th 2009, 02:33 AM

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:

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

**ozgunatalay** · November 29th 2009, 03:35 AM

Thank you for your solution.

**e^(i*pi)** · November 29th 2009, 03:37 AM

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

**ozgunatalay** · November 29th 2009, 03:48 AM

I am learning about this new. Sometimes when I'm having difficulty solving questions. Thanks for your suggestions.

**HallsofIvy** · November 29th 2009, 05:02 AM

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.

**ozgunatalay** · November 29th 2009, 05:25 AM

I looked again and I ask questions, I reached a different result. If I've made a mistake, correct my mistakes.

**ozgunatalay** · November 29th 2009, 09:35 AM

Is there anyone who can check if I did it right?

**hjortur** · November 29th 2009, 09:53 AM

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:

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

is supposed to be:

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

**ozgunatalay** · November 29th 2009, 10:51 AM

Where am I doing wrong?

**hjortur** · November 29th 2009, 10:54 AM

Nowhere. Your answer is the same as the corrected one.

**ozgunatalay** · November 29th 2009, 11:10 AM

**hjortur** · November 29th 2009, 11:41 AM

Sorry

, I didn't check out e^(i*pi)'s answer good enough.

But your answer is correct.

**ozgunatalay** · November 29th 2009, 12:26 PM

Thank you for your interest in question

Math Help - Derivatives of exponential functions

LinkBack

Thread Tools

Display

Derivatives of exponential functions

Similar Math Help Forum Discussions

Derivatives of Exponential Functions

Derivatives of Exponential Functions

derivatives exponential functions

Determining derivatives of exponential functions.

Derivatives Of inverse functions; Derivatives and integrals involving exponential fun

Search Tags