Answer
(a) \[\frac{\partial z}{\partial x}=f'(x)g(y)\]
\[\frac{\partial z}{\partial y}=f(x)g'(y)\]
(b) \[\frac{\partial z}{\partial x}=y\frac{\partial f}{\partial x}\]
\[\frac{\partial z}{\partial y}=x\frac{\partial f}{\partial y}\]
(c) \[\frac{\partial z}{\partial x}=\frac{1}{y}\frac{\partial f}{\partial x}\]
\[\frac{\partial z}{\partial y}=\left(-\frac{x}{y^2}\right)\frac{\partial f}{\partial y}\]
Work Step by Step
(a) \[z=f(x)g(y)\]
Differentiate $z$ with repsect to $x$ treating $y$ as constant.
\[\frac{\partial z}{\partial x}=\frac{df}{dx}g(y)\]
\[\frac{\partial z}{\partial x}=f'(x)g(y)\]
Differentiate $z$ with repsect to $y$ treating $x$ as constant.
\[\frac{\partial z}{\partial y}=f(x)\frac{dg}{dy}\]
\[\frac{\partial z}{\partial y}=f(x)g'(y)\]
(b) \[z=f(xy)\]
Differentiate $z$ with repsect to $x$ treating $y$ as constant using chain rule:
\[\frac{\partial z}{\partial x}=\left(\frac{\partial f}{\partial x}\right)y\frac{d}{dx}(x)\]
\[\frac{\partial z}{\partial x}=\left(\frac{\partial f}{\partial x}\right)y(1)\]
\[\frac{\partial z}{\partial x}=y\frac{\partial f}{\partial x}\]
Differentiate $z$ with repsect to $y$ treating $x$ as constant using chain rule:
\[\frac{\partial z}{\partial y}=\left(\frac{\partial f}{\partial y}\right)x\frac{d}{dy}(y)\]
\[\frac{\partial z}{\partial y}=\left(\frac{\partial f}{\partial y}\right)x(1)\]
\[\frac{\partial z}{\partial y}=x\frac{\partial f}{\partial y}\]
(c) \[z=f\left(\frac{x}{y}\right)\]
Differentiate $z$ with repsect to $x$ treating $y$ as constant using chain rule:
\[\frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}\cdot\frac{\partial}{\partial x}\left(\frac{x}{y}\right)\]
\[\frac{\partial z}{\partial x}=\frac{1}{y}\frac{\partial f}{\partial x}\]
Differentiate $z$ with repsect to $y$ treating $x$ as constant using chain rule:
\[\frac{\partial z}{\partial y}=\frac{\partial f}{\partial y}\cdot\frac{\partial}{\partial y}\left(\frac{x}{y}\right)\]
\[\frac{\partial z}{\partial y}=\left(-\frac{x}{y^2}\right)\frac{\partial f}{\partial y}\]
