# Differentiation of exponential

• Oct 9th 2008, 12:51 PM
Agent
Differentiation of exponential
Hey, I'm reviewing calculus (its been sometime since I took it) and I'm trying to do some examples from the book but I'm pretty sure I'm doing it wrong. :)
Heres the first one:

• Oct 9th 2008, 01:04 PM
Jhevon
Quote:

Originally Posted by Agent
Hey, I'm reviewing calculus (its been sometime since I took it) and I'm trying to do some examples from the book but I'm pretty sure I'm doing it wrong. :)
Heres the first one:

you either have to use the chain rule here, or distribute the power and use the product rule.

you seem to be attempting the chain rule. so i will ask you this, what is the derivative of $te^{-2t}$
• Oct 9th 2008, 03:48 PM
Agent
The exponential confuses me but I think its,

$
te^{-2t} * -2
$
• Oct 9th 2008, 03:54 PM
Jhevon
Quote:

Originally Posted by Agent
The exponential confuses me but I think its,

$
te^{-2t} * -2
$

well, it's not. you have a product of functions here, you need to use the product rule: $\frac d{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
• Oct 9th 2008, 04:03 PM
Agent
so can we say, f(x) = te and g(x) = -2t?
• Oct 9th 2008, 04:05 PM
Jhevon
Quote:

Originally Posted by Agent
so can we say, f(x) = te and g(x) = -2t?

no. $f(t) = t$ is one function, $g(t) = e^{-2t}$ is the other.

making $-2t$ as one function makes no sense, right? since you are saying there is some power without a base just floating around magically
• Oct 9th 2008, 04:17 PM
Agent
ok so if I'm following correctly,

$te^{-2t}$

$te^{-2t} * 2 + 1 * e^{-2t}$
• Oct 9th 2008, 04:24 PM
Jhevon
Quote:

Originally Posted by Agent
ok so if I'm following correctly,

$te^{-2t}$

$te^{-2t} * 2 + 1 * e^{-2t}$

almost. the derivative of $e^{-2t}$ is $-2e^{-2t}$. with that said,

$\frac d{dt}\bigg( t e^{-2t}\bigg) = -2te^{-2t} + e^{-2t} = e^{-2t}(1 - 2t)$

now, can you do the original problem?
• Oct 9th 2008, 04:50 PM
Agent
Quote:

Originally Posted by Jhevon
almost. the derivative of $e^{-2t}$ is $-2e^{-2t}$. with that said,

$\frac d{dt}\bigg( t e^{-2t}\bigg) = -2te^{-2t} + e^{-2t} = e^{-2t}(1 - 2t)$

now, can you do the original problem?

Awesome. I got it now. Thank You
• Oct 9th 2008, 04:57 PM
Jhevon
Quote:

Originally Posted by Agent
Awesome. I got it now. Thank You

do you have the answer in the back of your text or something? if not, you mind posting your solution?