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Finding the derivative of an equation with an exponent

  • Thread starter fk378
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1. The problem statement, all variables and given/known data
Find the local max and min
f(x)=xe^x



2. Relevant equations
first find the derivative of xe^x and then set it to 0



3. 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.
 
to find the derivative you need to use the product rule:

(uv)'=uv'+vu'

and from there you can find the critical points.
 

HallsofIvy

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(xex)'= (x)' ex+ x(ex)'

Now, are you saying that you do not know the derivative of ex?
 
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(xex)'= (x)' ex+ x(ex)'

Now, are you saying that you do not know the derivative of ex?

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

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

rock.freak667

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

[tex]f(x)=xe^x[/tex]

Take natural log of both sides:
[tex]ln(f(x))=ln(x)+x[/tex]

Use the chain rule to find the derivitive of the left side, and then differentiate the right:
[tex]\frac{f'(x)}{f(x)}=\frac{1}{x}+1[/tex]

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

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

nrqed

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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 [itex] e^x=0 [/itex] and is there any x for which [itex] (1+x) =0 [/itex] ?
 

HallsofIvy

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