Q. 35 - Calculus

Question

Find the first-order partial derivatives for the functions in Exercises 27–36.  

$f\left(x,y,z\right)=\frac{x{y}^2}{x+z}$

Step-by-Step Solution

Answer

The first-order partial derivatives are $\frac{\partial f}{\partial x}=\frac{z{y}^2}{{\left(x+z\right)}^2}, \frac{\partial f}{\partial y}=\frac{2xy}{x+z}, \frac{\partial f}{\partial z}=-\frac{x{y}^2}{{\left(x+z\right)}^2}.$

Step 1: Given

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

Step 2: To find

We have to find the first-order partial derivatives.  

Step 3: Calculation

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