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Second partial derivative of v=e^(x*e^y)

  • Thread starter sdoyle1
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  • #1
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Homework Statement


Find the second partial derivative of v=e^(x*e^y)


Homework Equations



I know that I need to find Vx and Vy first and then the second partial derivative would be Vxx, Vyy, Vxy.

The Attempt at a Solution



I'm really confused on how to find Vx or Vy
Vx= the derivative with regards to x, if y is a constant
so would it be Vx=e^(x*e^y) * (e^y)?
Any help would be great
 

Answers and Replies

  • #2
gb7nash
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Find the second partial derivative of v=e^(x*e^y)

I know that I need to find Vx and Vy first and then the second partial derivative would be Vxx, Vyy, Vxy.
You're forgetting one.

Vx=e^(x*e^y) * (e^y)
That's correct. What's Vy?
 
  • #3
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Would Vy= e^(x*e^y) * (xe^y)
 
  • #4
gb7nash
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Would Vy= e^(x*e^y) * (xe^y)
Correct. Now you need to take the second partial derivatives.
 
  • #5
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Vxx would be the second derivative with respect to x but keeping y as a constant. This is where I get confused. Would it be:

Vxx=e^(xye^y)(ye^y)
= e^(xy^2e^y)
 
  • #6
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Wait, you would add the exponents, not multiply them.
So Vx= e^((xe^y)+y)
Vxx= e^(xe^y+y) * (e^y)
 
  • #7
gb7nash
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  • #8
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Ok then Vyy= xe^((xe^y)+y) * (x(e^y) +1)

Then Vxy= e^((xe^y)+y) * ((xe^y)+1) and Vyx is the same as Vxy

As an aside, how would I integrate t(t-1)^1/2 ?
 
  • #9
gb7nash
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Ok then Vyy= xe^((xe^y)+y) * (x(e^y) +1)

Then Vxy= e^((xe^y)+y) * ((xe^y)+1) and Vyx is the same as Vxy
Correct.

As an aside, how would I integrate t(t-1)^1/2 ?
Try integration by parts.
 
  • #10
SammyS
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As an aside, how would I integrate t(t-1)^1/2 ?
Use the substitution: u = t-1 .
 
  • #11
gb7nash
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Good call. Both methods should work, but this is much simpler to do.
 

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