Finding the derivative of an equation with an exponent

[fk378]

Homework Statement

Find the local max and min
f(x)=xe^x

Homework Equations

first find the derivative of xe^x and then set it to 0

The Attempt at a Solution

I understand how to approach the problem but I don't know how to break apart the original equation since there is an exponent there. I know if I ln both sides I can get rid of the e and therefore bring down the x exponent, but ln(0) does not exist.

 

[bob1182006]

to find the derivative you need to use the product rule:

(uv)'=uv'+vu'

and from there you can find the critical points.

 

[HallsofIvy]

(xe^x)'= (x)' e^x+ x(e^x)'

Now, are you saying that you do not know the derivative of e^x?

 

[fk378]

(xe^x)'= (x)' e^x+ x(e^x)'

Now, are you saying that you do not know the derivative of e^x?

xe^x
f'(x)= (e^x) (x)(e^x)
to find the critical points:
(e^x) (x)(e^x)=x(e^2x) = 0

Now how do I solve for x when there is an x in the exponent?

 

[nrqed]

xe^x
f'(x)= (e^x) (x)(e^x)
to find the critical points:
(e^x) (x)(e^x)=x(e^2x) = 0

Now how do I solve for x when there is an x in the exponent?

You made a mistake. You must have a *sum* of two terms, not a product.
So you must solve $e^x + x e^x =0$ Factor out the exponential and find when this will be equal to zero.

 

[fk378]

You made a mistake. You must have a *sum* of two terms, not a product.
So you must solve $e^x + x e^x =0$ Factor out the exponential and find when this will be equal to zero.

So then I have e^x (1+x) = 0. How do I factor out the x when there is one in the exponent? I know I can take the natural log but can I do that for the 0 as well? but ln(0) does not exist.

 

[rock.freak667]

So then I have e^x (1+x) = 0. How do I factor out the x when there is one in the exponent? I know I can take the natural log but can I do that for the 0 as well? but ln(0) does not exist.

No need to take ln...if you know that e^x is never equal to zero for all values of x, then you can simply divide by e^x

 

[Saladsamurai]

No need to take ln...if you know that e^x is never equal to zero for all values of x, then you can simply divide by e^x

You have a product with two factors; only one of them has to equal zero in order for the entire product to equal zero.

 

[atqamar]

Here is another way to solve for the derivative. It is similar to the product rule, but I like it's elegance. And it has to do with the fact that $$\frac{d}{dx}ln(x)=\frac{1}{x}$$.

$$f(x)=xe^x$$

Take natural log of both sides:
$$ln(f(x))=ln(x)+x$$

Use the chain rule to find the derivative of the left side, and then differentiate the right:
$$\frac{f'(x)}{f(x)}=\frac{1}{x}+1$$

Since $$f(x)$$ is given above, multiply both sides by it and you end up with:
$$f'(x)=e^x+xe^x$$

Now when you set the derivative to $$0$$, you factor out the $$e^x$$ and use the zero product rule along with some function analysis to get your solution. Remember the analysis!
$$0=e^x(x+1)$$.

 

[nrqed]

So then I have e^x (1+x) = 0. How do I factor out the x when there is one in the exponent? I know I can take the natural log but can I do that for the 0 as well? but ln(0) does not exist.

You caNOT isolate x in that expression. So what youhave to do is to say the following: the product is zero if either of the two factors is zero. So you turn it into two separate problems. SI there any x for which $e^x=0$ and is there any x for which $(1+x) =0$ ?

 

[HallsofIvy]

(xe^x)'= (x)' e^x+ x(e^x)'

xe^x
f'(x)= (e^x) (x)(e^x)
to find the critical points:
(e^x) (x)(e^x)=x(e^2x) = 0

Now how do I solve for x when there is an x in the exponent?

Are you just ignoring what you are told? Use the product rule.
Didn't you see the "+" in my formula above?

 

[fk378]

(xe^x)'= (x)' e^x+ x(e^x)'

Are you just ignoring what you are told? Use the product rule.
Didn't you see the "+" in my formula above?

Yes I did see it and if you saw the other responses that followed you would also see that I fixed the mistake.