Derivatives of exponential functions

**e^(i*pi)** · November 29th 2009, 03:33 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, 04:35 AM

Thank you for your solution.

**e^(i*pi)** · November 29th 2009, 04:37 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, 04:48 AM

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

**HallsofIvy** · November 29th 2009, 06:02 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, 06:25 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, 10:35 AM

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

**hjortur** · November 29th 2009, 10:53 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, 11:51 AM

Where am I doing wrong?

**hjortur** · November 29th 2009, 11:54 AM

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

**ozgunatalay** · November 29th 2009, 12:10 PM

**hjortur** · November 29th 2009, 12:41 PM

Sorry

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

But your answer is correct.

**ozgunatalay** · November 29th 2009, 01:26 PM

Thank you for your interest in question

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