A fundamental theorem of calculus problem

Question

I'm asked to prove that $N_x - M_y = f_3(x, y, z)$, but:

$N(x, y, z) = \int_{x_0}^{x}f_3(u, y, z_0)du - \int_{z_0}^{z}f_1(x, y, u)du$

$M(x, y, z) = \int_{z_0}^{z}f_2(x, y, u)du$

So:

$$\begin{align} N_x &= \frac{\partial}{\partial x}(\int_{x_0}^{x}f_3(u, y, z_0)du - \int_{z_0}^{z}f_1(x, y, u)du) \\ &= f_3(x, y, z_0) \cdot \frac{\partial}{\partial x}(x) - f_1(x, y, z) \cdot \frac{\partial}{\partial y}(z) \\ &= f_3(x, y, z_0) \end{align}$$

$$\begin{align} M_y &= \frac{\partial}{\partial y}(\int_{z_0}^{z}f_2(x, y, u)du) \\ &= f_2(x, y, z) \cdot \frac{\partial}{\partial y}(z) \\ &= 0 \end{align}$$

Thus:

$N_x - M_y = f_3(x, y, z_0) \neq f_3(x, y, z)$

So, can I automatically say that the exercise is wrong because what they are asking me to prove is false or is there something I didn't see or is there another method to prove this?

Answer 1

The derivative $$ \frac{\partial}{\partial x}\int_{z_0}^{z}f_1(x, y, u)\,du $$ is wrong in your solution. It equals $$ \int_{z_0}^z \frac{\partial f_1}{\partial x}(x,y,u)\,du. $$ Likewise when you do $M_y$.