Question

Consider the function \(f(x) := exp(x+a)exp(-x)\)
a) Calculate the derivative \(f^{'}(x)\) by careful application of the product rule and the chain rule. Do not use the functional equation of exp.

Answer 1

First, tell me what the derivative of \(\exp(x)\) is.

Answer 2

hmmm exp(x) = \[\frac{d}{dx} \exp(x)\] possible ?

Answer 3

It is possible. In fact, it's very simple. \[\frac{d}{dx}\exp(x)=\exp(x)\]

Answer 4

ok..

Answer 5

Now, we need to apply the product rule and the chain rule. So that means \[\frac{d}{dx} (\exp(x-a)\exp(-x))=\left(\exp(x-a)\frac{d}{dx}\exp(-x)+\exp(-x)\frac{d}{dx}\exp(x-a)\right)\]
Can you solve it yourself from here?

Answer 6

i will try but to be honest i dont trust my self..

Answer 7

:)

Answer 8

Just write out your final solution here, and I'll check you.

Answer 9

ok ...

Answer 10

\[\exp(-2a)\frac{d}{dx}\] is it ?

Answer 11

I'm getting \[-\exp(x-a)\exp(-x)+\exp(-x)\exp(x-a)\]

Answer 12

But that doesn't seem right. One minute.

Answer 13

ok

Answer 14

I'm definitely still getting this when I use the product rule and chain rule\[−\exp(x−a)\exp(−x)+\exp(−x)\exp(x−a)=0\]

Answer 15

you should since \[\exp(x+a)\exp(-x)=\exp(x+a-x)=\exp(a)\]

which is a constant....the derivative of a constant is zero

Answer 16

Right, I was under the foolish impression that \[\frac{d}{dx}\exp(a)=\exp(a)\]

Answer 17

can i write both what Zarkon wrote and you wrote George?

Answer 18

the instructions say "by careful application of the product rule and the chain rule"

Answer 19

I used neither

Answer 20

ok..

Answer 21

So don't use what he wrote.

Answer 22

also i write as a conclusion this right ? \[ \frac{d}{dx}\exp(a)=\exp(a)\]

Answer 23

and what you wrote before..

Answer 24

No. That statement is completely wrong. It should read \[\frac{d}{dx}\exp(a)=0\]However, that's not your solution. You want to use what I had written previously.

Answer 25

aah ok.. thanks George

Answer 26

You're welcome.