Derivatives of e^x: Solutions and Explanations

[Elihu5991]

Homework Statement

Find the derivative of these e^x functions [paraphrased]

Homework Equations

$x^{2} e^{x}$

The Attempt at a Solution

$2x e^{x}$

This is how I believe it to be correct from my understanding. It does feel wrong, with the answers agreeing with the feeling.

 

[trollcast]

Homework Statement

Find the derivative of these e^x functions [paraphrased]

Homework Equations

$x^{2} e^{x}$

The Attempt at a Solution

$2x e^{x}$

This is how I believe it to be correct from my understanding. It does feel wrong, with the answers agreeing with the feeling.

You need to use the product rule to differentiate that.

$$\frac{d}{dx}(u\cdot v)=u(\frac{dv}{dx})+ v(\frac{du}{dx})$$

 

[BruceW]

the function $x^2e^x$ is a product of two functions. So what rule do you need to use, to take the derivative?

 

[Elihu5991]

Can't believe I didn't notice the use of the product rule. As I was progressing through the question, I wasn't sure of one new thing: is two $e^{x}$ equal to $e^{x}$ or $2e^{x}$? I reckon it's the latter, but the answers are in the simplest form and doesn't seem to fit; unless I've made another minute error.

 

[Karnage1993]

$e^x + e^x = 2e^x$ if that is what you're asking. Pretty standard stuff...

 

[mtayab1994]

Can't believe I didn't notice the use of the product rule. As I was progressing through the question, I wasn't sure of one new thing: is two $e^{x}$ equal to $e^{x}$ or $2e^{x}$? I reckon it's the latter, but the answers are in the simplest form and doesn't seem to fit; unless I've made another minute error.

Well try to look at x^2e^x as u=x^2 and v=e^x

so (uv)'=u'v+v'u. Now just compute that and you will find your derivative.

You should get 2xe^x+x^2e^x=x(2e^x+xe^x)=xe^x(x+2)

 

[Elihu5991]

Yeah that's what I thought. But I had to check as I was getting problems. It would be my working-out. We were doing some different than standard stuff with Euler's Number today, so I wondered if it applied to this situation as well.

 

[Elihu5991]

Well try to look at x^2e^x as u=x^2 and v=e^x

so (uv)'=u'v+v'u. Now just compute that and you will find your derivative.

You should get 2xe^x+x^2e^x=x(2e^x+xe^x)=xe^x(x+2)

Did you differently simplify it? The answer in my book: $xe^{x}(2+x)$

 

[mtayab1994]

Did you differently simplify it? The answer in my book: $xe^{x}(2+x)$

No it's the same look at my post.

 

[Elihu5991]

Oh whoops, I somehow didn't see that segment ...

 

[Elihu5991]

Thankyou everyone for your help.

P.S How do I mark this topic as [SOLVED] ?

 

[trollcast]

Thankyou everyone for your help.

P.S How do I mark this topic as [SOLVED] ?

There's no solved prefixes on the forums.

 

[Elihu5991]

Oh ok. Well, thank you!

P.S Turns out that our teacher accidentally jumped ahead in the course. Explaining why I seemed to ask silly questions, that I'm now able to since we had the grounding today.