Q. 25 - Calculus

Question

Use the definition of the partial derivative to find the partial derivatives specified in Exercises 23–26.  

$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\text{ and} \frac{\partial f}{\partial z} \text{when} f\left(x, y, z\right)=\frac{x{y}^2}{z}$

Step-by-Step Solution

Answer

Required partial derivatives are: $\frac{\partial f}{\partial x}=\frac{y^2}{z}, \frac{\partial f}{\partial y}=\frac{2xy}{x} \text{ and} \frac{\partial f}{\partial z}=\frac{-x{y}^2}{z^2}$

Step 1: Given

The given function is: $f\left(x, y, z\right)=\frac{x{y}^2}{z}$.

Step 2: To find

We have to find $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\text{ and} \frac{\partial f}{\partial z}.$

Step 3: Calculation

Differentiate both sides with respect to x, y, and z respectively.

$$\begin{gathered} f\left(x, y, z\right)=\frac{x{y}^2}{z} \\ \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x{y}^2}{z}\right) \\ \frac{\partial f}{\partial x}=\frac{y^2}{z} \\ f\left(x, y, z\right)=\frac{x{y}^2}{z} \\ \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(\frac{x{y}^2}{z}\right) \\ \frac{\partial f}{\partial y}=\frac{2xy}{x} \\ f\left(x, y, z\right)=\frac{x{y}^2}{z} \\ \frac{\partial f}{\partial z}=\frac{\partial }{\partial z}\left(\frac{x{y}^2}{z}\right) \\ \frac{\partial f}{\partial z}=\frac{\partial }{\partial z}\left(x{y}^2{z}^{-1}\right) \\ \frac{\partial f}{\partial z}=x{y}^2\left(-1\right)z^{-2} \\ \frac{\partial f}{\partial z}=\frac{-x{y}^2}{z^2} \\ \text{Hence}, \frac{\partial f}{\partial x}=\frac{y^2}{z}, \frac{\partial f}{\partial y}=\frac{2xy}{x} \text{ and} \frac{\partial f}{\partial z}=\frac{-x{y}^2}{z^2} \end{gathered}$$