1. ## 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:

2. 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 $te^{-2t}$

3. The exponential confuses me but I think its,

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

4. 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: $\frac d{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

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

6. 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

7. ok so if I'm following correctly,

$te^{-2t}$

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

8. 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?

9. 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

10. 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?