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

• Jan 6th 2010, 03:57 AM
Blacklaw
Derivate 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 AM
dedust
Quote:

Originally Posted by Blacklaw
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.

$\displaystyle f(x) = g(x)h(x)$, then $\displaystyle f'(x) = g'(x)h(x) + g(x)h'(x)$

take $\displaystyle g(x) = x^2$ and $\displaystyle h(x) = e^{-3x}$
• Jan 6th 2010, 04:12 AM
Blacklaw
Quote:

Originally Posted by dedust
$\displaystyle f(x) = g(x)h(x)$, then $\displaystyle f'(x) = g'(x)h(x) + g(x)h'(x)$

take $\displaystyle g(x) = x^2$ and $\displaystyle h(x) = e^{-3x}$

I get $\displaystyle 2x*e^{-3x}+x^2*e^{-3x}$

But I know for a fact that the correct answer is $\displaystyle 2x*e^{-3x}+x^2*e^{-3x}*(-3)$

Please explain how I get the (-3)?
• Jan 6th 2010, 04:16 AM
CaptainBlack
Quote:

Originally Posted by Blacklaw
I get $\displaystyle 2x*e^{-3x}+x^2*e^{-3x}$

But I know for a fact that the correct answer is $\displaystyle 2x*e^{-3x}+x^2*e^{-3x}*(-3)$

Please explain how I get the (-3)?

$\displaystyle \frac{d}{dx}e^{-3x}=-3e^{-3x}$

You did not know the answer for a fact if it was what was in the back of the book. You know it for a fact if you know why it is so and now you should do.

CB
• Jan 6th 2010, 04:31 AM
Blacklaw
Quote:

Originally Posted by CaptainBlack
$\displaystyle \frac{d}{dx}e^{-3x}=-3e^{-3x}$

You did not know the answer for a fact if it was what was in the back of the book. You know it for a fact if you know why it is so and now you should do.

CB

Does it always go like that? I thought the derivative of $\displaystyle e^x$ was $\displaystyle e^x$?

Thank you for you help, I truly appreciate it.
• Jan 6th 2010, 05:24 AM
tom@ballooncalculus
Quote:

Originally Posted by Blacklaw
Does it always go like that?

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 AM
Blacklaw
I think I get it now.. Thank you very much for helping me out.
• Jan 6th 2010, 09:13 AM
rshekhar.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 AM
CaptainBlack
Quote:

Originally Posted by rshekhar.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.

What you label as "by chain riue" is mainly product rule and to a lesser extent chain rule

CB
• Jan 6th 2010, 09:26 AM
CaptainBlack
Quote:

Originally Posted by Blacklaw
Does it always go like that? I thought the derivative of $\displaystyle e^x$ was $\displaystyle e^x$?

Thank you for you help, I truly appreciate it.

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