Differentiation of exponential

**Agent** · October 9th 2008, 11:51 AM

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:

**Jhevon** · October 9th 2008, 12:04 PM

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 $te^{-2t}$

**Agent** · October 9th 2008, 02:48 PM

The exponential confuses me but I think its,

$<br /> te^{-2t} * -2<br />$

**Jhevon** · October 9th 2008, 02:54 PM

Originally Posted by Agent

The exponential confuses me but I think its,

$<br /> te^{-2t} * -2<br />$

well, it's not. you have a product of functions here, you need to use the product rule: $\frac d{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

**Agent** · October 9th 2008, 03:03 PM

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

**Jhevon** · October 9th 2008, 03:05 PM

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

**Agent** · October 9th 2008, 03:17 PM

ok so if I'm following correctly,

$te^{-2t}$

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

**Jhevon** · October 9th 2008, 03:24 PM

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?

**Agent** · October 9th 2008, 03:50 PM

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

**Jhevon** · October 9th 2008, 03:57 PM

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?

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