Answer
$$\frac{\partial z}{\partial x}=5\cos5x\cos5y$$
$$\frac{\partial z}{\partial y}=-5\sin5x\sin5y$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(\sin5x\cos5y)=\cos5y\frac{\partial }{\partial x}(\sin5x)=\cos5y\cos5x\frac{\partial }{\partial x}(5x)=\cos5y\cos5x\cdot5=5\cos5x\cos5y$$
Here is $y$ treated as a constant, so $\cos5y$ is treated as a constant as well. We used chain rule to find partial derivative $\sin5x.$
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(\sin5x\cos5y)=\sin5x\frac{\partial }{\partial y}(\cos5y)=\sin5x\cdot(-\sin5y)\frac{\partial }{\partial y}(5y)=-\sin5x\sin5y\cdot5=-5\sin5x\sin5y$$
We treated $x$ here as a constant, as we did with $\sin5x$. We used chain rule to find partial derivative $\frac{\partial }{\partial y}(\cos5y).$
