OpenStudy (anonymous):

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.

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (anonymous):

ok..

5 years ago
OpenStudy (kinggeorge):

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?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

:)

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (anonymous):

ok ...

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (kinggeorge):

But that doesn't seem right. One minute.

5 years ago
OpenStudy (anonymous):

ok

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (zarkon):

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

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (zarkon):

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

5 years ago
OpenStudy (zarkon):

I used neither

5 years ago
OpenStudy (anonymous):

ok..

5 years ago
OpenStudy (kinggeorge):

So don't use what he wrote.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

and what you wrote before..

5 years ago
OpenStudy (kinggeorge):

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.

5 years ago
OpenStudy (anonymous):

aah ok.. thanks George

5 years ago
OpenStudy (kinggeorge):

You're welcome.

5 years ago
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