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Second partial derivatives

  • Thread starter fk378
  • Start date
367
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1. Homework Statement
Find the second partial derivatives.
z= x/(x+y)



3. The Attempt at a Solution
I solved the correct df/dx, d^2f/dx^2, df/dy, and d^2f/dy^2, however I can't seem to get the correct answer for d^2f/dydx and d^2f/dxdy.

My df/dx is y/(x+y)^2 which I changed to y((x+y)^-2)

Differentiating with respect to y, using the product rule: (1)(x+y)^-2 + y(-2)((x+y)^-3) = (x+y)^-2 - 2y((x+y)^-3)

My df/dy is -x((x+y)^-2)

Differentiating with respect to x, using the product rule: (-1)(x+y)^-2 + (-x)(-2)((x+y)^-3) = -(x+y)^-2 + 2x((x+y)^-3)

I know they aren't right b/c 1) they don't equal each other (Clairaut's Thm.) and 2) it doesn't match the book's answer =)

Can anyone catch the mistake I'm making? Thank you.
 
Last edited:

Answers and Replies

22
0
I think you're right. You seem to have done everything ok. Just try to simplifiy both result solving the newton binomials. Probably everything will reduce and you'll have your right awnser.
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367
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Hm, I don't think it works out. I just tried to simplify it and I still didn't get the right answer. Also, my d^2f/dydx answer has a y variable in it, while my d^2f/dxdy answer has an x variable in it. And simplifying either expression doesn't get rid of them.

The correct answer is supposed to be (x-y)((x+y)^-3)
 
olgranpappy
Homework Helper
1,271
3
you can simplify df/dx to
[tex]
\frac{y}{(x+y)^2}
[/tex]
and you can simplify df/dy to
[tex]
\frac{-x}{(x+y)^2}
[/tex]
Taking d/dx on the above gives
[tex]
(-1)(x+y)^{(-2)}+2x/(x+y)^3
=\frac{x-y}{(x+y)^3}\;.
[/tex]
You can also show that d^2f/dydx is the same
 
367
0
Isn't the denominator supposed to be [g(x)]^2 so why isn't the denominator (x+y)^4?

And, can you see what I did wrong when I used the product rule?
 
olgranpappy
Homework Helper
1,271
3
you didn't do anything wrong. df/dx and df/dy are *not* "supposed to equal each other". It is
[tex]
\frac{d^2 f}{dxdy}
[/tex]
which is supposed to equal
[tex]
\frac{d^2 f}{dydx}
[/tex]

The expression you gave for d^2f/dxdy is
[tex]
-\frac{1}{(x+y)^2} + \frac{2x}{(x+y)^3}
[/tex]
which can be simplified to the book's answer using algebra (multiply the first term by (x+y)/(x+y) and then the terms have the same denominators and you can add the numerators)
 
367
0
I can't see your latex image for some reason--it says it is invalid. Also, my answers are not the same. One has a y variable, the other has an x variable. Also there are different signs within each expression.
 
olgranpappy
Homework Helper
1,271
3
hmm... I don't know why you cant see the latex... maybe try reloading.
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
No, there is no mistake- that's what everyone is trying to tell you. Yes, If you had done this using the "quotient" rule, the denominator would have (x+ y)4. But the numerator would have a factor of (x+ y). Try it!

1. Homework Statement
Find the second partial derivatives.
z= x/(x+y)



3. The Attempt at a Solution
I solved the correct df/dx, d^2f/dx^2, df/dy, and d^2f/dy^2, however I can't seem to get the correct answer for d^2f/dydx and d^2f/dxdy.

My df/dx is y/(x+y)^2 which I changed to y((x+y)^-2)

Differentiating with respect to y, using the product rule: (1)(x+y)^-2 + y(-2)((x+y)^-3) = (x+y)^-2 - 2y((x+y)^-3)
So [tex]\frac{\partial^2 f}{\partial x\partial y}= \frac{1}{(x+y)^2}+ \frac{-2y}{(x+y)^3}[/tex]
[tex]= \frac{x+ y}{(x+y)^3}+ \frac{-2y}{(x+y)^3}= \frac{x- y}{(x+y)^3}[/tex]

My df/dy is -x((x+y)^-2)

Differentiating with respect to x, using the product rule: (-1)(x+y)^-2 + (-x)(-2)((x+y)^-3) = -(x+y)^-2 + 2x((x+y)^-3)
So [tex]\frac{\partial^2 f}{\partial y\partial x}= \frac{-1}{(x+ y)^2}+ \frac{2x}{(x+y)^3}[/tex]
[tex]= \frac{-(x+y)}{(x+y)^3}+ \frac{2x}{(x+y)^3}= \frac{x- y}{(x+y)^3}[/tex]

I know they aren't right b/c 1) they don't equal each other (Clairaut's Thm.) and 2) it doesn't match the book's answer =)

Can anyone catch the mistake I'm making? Thank you.
Looks to me like they are exactly the same!
 
367
0
I see it!! Thank you all =)

I did not catch the multiplication of the expression by (x+y)/(x+y)
 

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