Hi guys

I've been trying to solve this for a little while now but I feel kinda lost and I thought you could maybe help me out.

so

f(x)=x^2*e^(-3x)

f '(x)=?

Thank you.

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- Jan 6th 2010, 03:57 AMBlacklawDerivate f(x)=x^2*e*(-3x)
Hi guys

I've been trying to solve this for a little while now but I feel kinda lost and I thought you could maybe help me out.

so

f(x)=x^2*e^(-3x)

f '(x)=?

Thank you. - Jan 6th 2010, 04:01 AMdedust
- Jan 6th 2010, 04:12 AMBlacklaw
- Jan 6th 2010, 04:16 AMCaptainBlack
- Jan 6th 2010, 04:31 AMBlacklaw
- Jan 6th 2010, 05:24 AMtom@ballooncalculus
That's what the chain rule says.

Just in case a picture helps...

http://www.ballooncalculus.org/asy/chain.png

is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

In this case...

http://www.ballooncalculus.org/asy/d...expMinusU1.png

PS:

Just in case another picture doesn't hurt (and in view of comments below)...

Wrapping the chain rule inside the (legs-uncrossed version of) ...

http://www.ballooncalculus.org/asy/prod.png

... the product rule, we get...

http://www.ballooncalculus.org/asy/d...expMinusU2.png

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jan 6th 2010, 05:34 AMBlacklaw
I think I get it now.. Thank you very much for helping me out.

- Jan 6th 2010, 09:13 AMrshekhar.in
$\displaystyle f(x) = x^2 e^{-3x}$

$\displaystyle f'(x) = x^2 e^{-3x} (-3) + e^{-3x} 2x$ By chain rule

$\displaystyle f'(x) = x e^{-3x} [ -3x + 2 ] $

Hope this helps. - Jan 6th 2010, 09:23 AMCaptainBlack
- Jan 6th 2010, 09:26 AMCaptainBlack
The derivative of $\displaystyle e^x$ is $\displaystyle e^x$ but that is not what you have, what you should know is that the derivative of $\displaystyle e^{ax}$ is $\displaystyle ae^{ax}$ by the chain rule if you must, but you should know it because it will occur so often.

CB