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.