# Second partial derivatives

## Homework Statement

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

## 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.

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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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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
you can simplify df/dx to
$$\frac{y}{(x+y)^2}$$
and you can simplify df/dy to
$$\frac{-x}{(x+y)^2}$$
Taking d/dx on the above gives
$$(-1)(x+y)^{(-2)}+2x/(x+y)^3 =\frac{x-y}{(x+y)^3}\;.$$
You can also show that d^2f/dydx is the same

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
you didn't do anything wrong. df/dx and df/dy are *not* "supposed to equal each other". It is
$$\frac{d^2 f}{dxdy}$$
which is supposed to equal
$$\frac{d^2 f}{dydx}$$

The expression you gave for d^2f/dxdy is
$$-\frac{1}{(x+y)^2} + \frac{2x}{(x+y)^3}$$
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)

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
hmm... I don't know why you cant see the latex... maybe try reloading.

HallsofIvy
Homework Helper
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!

## Homework Statement

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

## 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 $$\frac{\partial^2 f}{\partial x\partial y}= \frac{1}{(x+y)^2}+ \frac{-2y}{(x+y)^3}$$
$$= \frac{x+ y}{(x+y)^3}+ \frac{-2y}{(x+y)^3}= \frac{x- y}{(x+y)^3}$$

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)
So $$\frac{\partial^2 f}{\partial y\partial x}= \frac{-1}{(x+ y)^2}+ \frac{2x}{(x+y)^3}$$
$$= \frac{-(x+y)}{(x+y)^3}+ \frac{2x}{(x+y)^3}= \frac{x- y}{(x+y)^3}$$

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.
Looks to me like they are exactly the same!

I see it!! Thank you all =)

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