first and second order partial derivatives

**andrey** · December 30th 2011, 04:41 AM

Hi! I’m novice in the field and surely my question would be very simple for you. I’m now reading a textbook of Mathematics for biologists and try to understand the solved problems in the text. I have trouble with this one:
Find all partial derivatives up to second order of the function: z=(x+y)/(x-y)

The book says:

first and second order partial derivatives-derivatives.jpg

I am confused with ∂/∂y(∂z/∂x) and ∂/∂y(∂z/∂y). Trying to solve the problem, I get:

first and second order partial derivatives-my_results.jpg

I know that for the most functions of two variables ∂/∂y(∂z/∂x) = ∂/∂x(∂z/∂y) and probably the textbook isn’t wrong, so I would be grateful if you point where I am wrong.

Thank you very much in advance!

**alexmahone** · December 30th 2011, 05:50 AM

Originally Posted by andrey

Hi! I’m novice in the field and surely my question would be very simple for you. I’m now reading a textbook of Mathematics for biologists and try to understand the solved problems in the text. I have trouble with this one:
Find all partial derivatives up to second order of the function: z=(x+y)/(x-y)

The book says:

Click image for larger version.

Name: derivatives.JPG
Views: 10
Size: 35.1 KB
ID: 23164

I am confused with ∂/∂y(∂z/∂x) and ∂/∂y(∂z/∂y). Trying to solve the problem, I get:

Click image for larger version.

Name: my_results.JPG
Views: 12
Size: 21.0 KB
ID: 23163

I know that for the most functions of two variables ∂/∂y(∂z/∂x) = ∂/∂x(∂z/∂y) and probably the textbook isn’t wrong, so I would be grateful if you point where I am wrong.

Thank you very much in advance!

$\frac{\partial}{\partial{y}}(x-y)^2=-2(x-y)$

**MarceloFantini** · December 30th 2011, 05:51 AM

I believe the error is here:

$\frac{\partial}{\partial y}((x-y)^2) = -2(x-y)$

And not $2(x-y)$ as you used. Try working those steps.

**Soroban** · December 30th 2011, 07:08 AM

Hello, andrey!

$\text{Find all partial derivatives up to second order of the function: }\:z \:=\: \frac{x+y}{x-y}$

$\frac{\partial z}{\partial x} \;=\;\frac{(x-y)(1) - (x+y)(1)}{(x-y)^2} \;=\;\frac{-2y}{(x-y)^2} \;=\;-2y(x-y)^{-2}$

$\frac{\partial z}{\partial y} \;=\;\frac{(x-y)(1) - (x+y)(\text{-}1)}{(x-y)^2} \;=\;\frac{2x}{(x-y)^2} \;=\;2x(x-y)^{-2}$

$\frac{\partial^2z}{\partial x^2} \;=\;-2y(\text{-}2)(x-y)^{-3}(1) \;=\;\frac{4y}{(x-y)^3}$

$\frac{\partial^2z}{\partial y^2} \;=\;2x(\text{-}2)(x-y)^{-3}(\text{-}1) \;=\;\frac{4x}{(x-y)^3}$

$\frac{\partial^2z}{\partial y\partial x} \;=\;\frac{(x-y)^2(\text{-}2) - (\text{-}2y)(2)(x-y)(\text{-}1)}{(x-y)^4}$

. . . . . $=\;\frac{-2(x-y)\big[(x-y)+2y\big]}{(x-y)^4} \;=\; \frac{-2(x+y)}{(x-y)^3}$

**andrey** · December 30th 2011, 08:00 AM

Thank you so much for your replies!
The clarification that ∂/∂y((x-y)^2)=-2(x-y) was really important in my case.

Soroban, thank you very much for the detailed solution!

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