Q. 24 - Calculus

Question

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

$f_x\left(0,-3\right)\text{ and} f_y\left(0,-3\right) \text{when} f\left(x,y\right)=\frac{3x}{y^2}$

Step-by-Step Solution

Answer

Required partial derivatives are$f_x\left(0,-3\right)=\frac{1}{3} \text{and} f_y\left(0,-3\right)=0$

Step 1: Given

The given function is:  $f\left(x,y\right)=\frac{3x}{y^2}$.

Step 2: To find

We have to find $f_x\left(0,-3\right)\text{ and} f_y\left(0,-3\right)$.

Step 3: Calculation

Differentiate both sides with respect to x and y respectively and then plug $x = 0$ and $y=-3$.

$$\begin{gathered} f\left(x,y\right)=\frac{3x}{y^2} \\ {f}_x\left(x,y\right)=\frac{\partial }{\partial x}\left(\frac{3x}{y^2}\right) \\ {f}_x\left(x,y\right)=\frac{3}{y^2} \\ {f}_x\left(0,-3\right)=\frac{3}{{\left(-3\right)}^2} \\ {f}_x\left(0,-3\right)=\frac{3}{9} \\ {f}_x\left(0,-3\right)=\frac{1}{3} \\ f\left(x,y\right)=\frac{3x}{y^2} \\ {f}_y\left(x,y\right)=\frac{\partial }{\partial y}\left(\frac{3x}{y^2}\right) \\ {f}_y\left(x,y\right)=\frac{\partial }{\partial y}\left(3x{y}^{-2}\right) \\ {f}_y\left(x,y\right)=3x\left(-2\right)y^{-3} \\ {f}_y\left(x,y\right)=-\frac{6x}{y^3} \\ {f}_y\left(0,-3\right)=-\frac{6\left(0\right)}{{\left(-3\right)}^3} \\ {f}_y\left(0,-3\right)=0 \\ \text{Hence}, f_x\left(0,-3\right)=\frac{1}{3} \text{and} f_y\left(0,-3\right)=0 \end{gathered}$$