Im trying to find following first order partial derivative in the given point:

$$ f(x,y) = \frac{x-y}{x+y} \text{at} (2,-1)$$

Im sort of confused as to how to solve this. I have tried the quotiënt rule which gave me the following:

$$ (\frac{f}{g})' = \frac{f'*g - g'*f}{g^2} $$

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

with $x = 2$ and $y = -1$ i get the correct solution of $-4$. But when i apply the same method for the partial derivative to x i keep getting a wrong answer.

Am I coming at this the right way or not? All pointers would be very helpful, thanks.


We have $$f(x,y) = \frac{x-y}{x+y}$$ Thus, we have, $$\frac{\partial f}{\partial x} = \frac{(x+y)(1)-(x-y)(1)}{(x+y)^2} = \frac{2y}{(x+y)^2}$$ Substituting $x = 2, y=-1$, we get, $$\frac{\partial f}{\partial x} =-2$$ Similarly, $$\frac{\partial f}{\partial y} = \frac{(x+y)(-1)-(x-y)(1)}{(x+y)^2} =-\frac{2x}{(x+y)^2}$$ Substituting $x =2, y=-1$, we get, $$\frac{\partial f}{\partial y} = -4$$ And also note that, in general, $$\frac{\partial f}{\partial x} \neq \frac{\partial f}{\partial y}$$ Hope it helps.

  • $\begingroup$ I was thinking that this would be the solution. Only problem is I don't see how the quotient rule formula can change since $f$ and $g$ and their respective derivatives used in the formula should stay the same, no? I see that in the derivative to $x$, you multiply by $1$ twice In the derivative to $y$ you multiply by $1$ once and by $-1$ once. How does the sign change here? $\endgroup$ – TheAlPaca02 Dec 26 '16 at 12:01
  • 1
    $\begingroup$ We are taking the partial derivative of say, $x+y$ w.r.t. to x and we get the answer as 1. Similarly for y we get 1. But, when we take partial derivative of $x-y$, that is, $x+(-y)$ the partial derivative w.r.t to x is still 1 . But, w.r.t y , it is -1. That is the reason for the difference in the answers. $\endgroup$ – Rohan Dec 26 '16 at 12:05

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