Q. 26 - Calculus

Question

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

$\frac{\partial g}{\partial x} \text{and }\frac{\partial g}{\partial y} \text{when} g\left(x,y\right)=\frac{\sqrt{y}}{x+1}$.

Step-by-Step Solution

Answer

Required partial derivatives are $\frac{\partial g}{\partial x}=-\frac{\sqrt{y}}{{\left(x+1\right)}^2} \text{and} \frac{\partial g}{\partial y}=\frac{1}{2\left(x+1\right)\sqrt{y}}.$

Step 1: Given

The given function is: $g\left(x,y\right)=\frac{\sqrt{y}}{x+1}$.

Step 2: To find

We have to find $\frac{\partial g}{\partial x} \text{and }\frac{\partial g}{\partial y}\text{.}$

Step 3: Calculation

Differentiate both sides with respect to x and y respectively.

$$\begin{gathered} g\left(x,y\right)=\frac{\sqrt{y}}{x+1} \\ \frac{\partial g}{\partial x}=\frac{\partial }{\partial x}\left(\frac{\sqrt{y}}{x+1}\right) \\ \frac{\partial g}{\partial x}=\frac{\partial }{\partial x}\left(\sqrt{y}{\left(x+1\right)}^{-1}\right) \\ \frac{\partial g}{\partial x}=\sqrt{y}\left(-1\right){\left(x+1\right)}^{-2}\frac{\partial }{\partial x}\left(x+1\right) \\ \frac{\partial g}{\partial x}=\sqrt{y}\left(-1\right){\left(x+1\right)}^{-2}\left(1\right) \\ \frac{\partial g}{\partial x}=-\frac{\sqrt{y}}{{\left(x+1\right)}^2} \\ g\left(x,y\right)=\frac{\sqrt{y}}{x+1} \\ \frac{\partial g}{\partial y}=\frac{\partial }{\partial y}\left(\frac{\sqrt{y}}{x+1}\right) \\ \frac{\partial g}{\partial y}=\frac{\partial }{\partial y}\left(\frac{y^{{\displaystyle \frac{1}{2}}}}{x+1}\right) \\ \frac{\partial g}{\partial y}=\frac{1}{\left(x+1\right)}\frac{\partial }{\partial y}\left(y^{\frac{1}{2}}\right) \\ \frac{\partial g}{\partial y}=\frac{1}{\left(x+1\right)}\left(\frac{1}{2}\right)\left(y^{-\frac{1}{2}}\right) \\ \frac{\partial g}{\partial y}=\frac{1}{2\left(x+1\right)\sqrt{y}} \\ \text{Hence}, \frac{\partial g}{\partial x}=-\frac{\sqrt{y}}{{\left(x+1\right)}^2} \text{and} \frac{\partial g}{\partial y}=\frac{1}{2\left(x+1\right)\sqrt{y}}. \end{gathered}$$