Q. 72 - Calculus

Question

$$\begin{gathered} \mathrm{Let} z = f(x, y)g(x, y). \mathrm{Show} \mathrm{that} \frac{\partial z}{\partial x}= \frac{\partial f}{\partial x} g(x, y) + f(x, y) \frac{\partial g}{\partial x} \mathrm{and} \\ \frac{\partial z}{\partial y}= \frac{\partial f}{\partial y} g(x, y) + f(x, y) \frac{\partial g}{\partial y} \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \frac{\partial z}{\partial x}= \frac{\partial f}{\partial x} g(x, y) + f(x, y) \frac{\partial g}{\partial x} \\ \frac{\partial z}{\partial y}= \frac{\partial f}{\partial y} g(x, y) + f(x, y) \frac{\partial g}{\partial y} \end{gathered}$$

Step 1. Given information

A function, $z = f(x, y)g(x, y)$

Step 2. Proofs of given partial derivatives

$$\begin{gathered} \mathrm{LHS}=\frac{\partial z}{\partial x} \\ =\frac{\partial f(x,y)g(x,y)}{\partial x} \\ \mathrm{Using} \mathrm{product} \mathrm{rule} \mathrm{and} \mathrm{definition} \mathrm{of} \mathrm{partial} \mathrm{derivatives}, \\ =\frac{\partial f}{\partial x} g(x, y) + f(x, y) \frac{\partial g}{\partial x} \\ \mathrm{Similarly}, \mathrm{for} \mathrm{partial} \mathrm{derivative} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{y}, \mathrm{we} \mathrm{get}, \\ \mathrm{LHS}=\frac{\partial z}{\partial y} \\ =\frac{\partial f(x,y)g(x,y)}{\partial y} \\ \mathrm{Using} \mathrm{product} \mathrm{rule} \mathrm{and} \mathrm{definition} \mathrm{of} \mathrm{partial} \mathrm{derivatives}, \\ = \frac{\partial f}{\partial y} g(x, y) + f(x, y) \frac{\partial g}{\partial y} \\ =\mathrm{RHS} \end{gathered}$$