Applying chain rule to $f(x,y) = -h(x,-y)$

Question

Could someone check my work? Thanks!

$$-f_x(x,y) = \frac{\partial h(x,-y)}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial h(x,-y)}{\partial (-y)}\frac{\partial (-y)}{\partial x} = h_x(x,-y),$$

$$- f_y(x,y) = \frac{\partial h(x,-y)}{\partial x}\frac{\partial x}{\partial y} + \frac{\partial h(x,-y)}{\partial (-y)}\frac{\partial (-y)}{\partial y} = -h_{-y}(x,-y).$$

Therefore, we have $$f_x(x,y) = -h_x(x,-y) \qquad \text{and} \qquad f_y(x,y) = h_{-y}(x,-y).$$

However, some people get $$f_x(x,y) = -h_x(x,-y) \qquad \text{and} \qquad f_y(x,y) = h_{y}(x,-y).$$

Which one is correct? If the second one is correct, how do we get it?

Answer 1

I'd say

$$-f_y(x,y)=\frac{\partial h(x,-y)}{\partial y}=h_y(x,-y)\cdot\frac{d(-y)}{dy}=-h_y(x,-y)$$