Q. 30 - Calculus

Question

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

$g\left(x,y\right)=\frac{\ln \left(y^2+1\right)}{x}$

Step-by-Step Solution

Answer

The first-order partial derivatives are $\frac{\partial g}{\partial x}=-\frac{\ln \left(y^2+1\right)}{x^2} \text{and }\frac{\partial g}{\partial y}=\frac{2y}{x\left(y^2+1\right)}$

Step 1: Given

The given function is: $g\left(x,y\right)=\frac{\ln \left(y^2+1\right)}{x}$

Step 2: To find

We have to find the first-order partial derivatives.  

Step 3: Calculation

$$\begin{gathered} g\left(x,y\right)=\frac{\ln \left(y^2+1\right)}{x} \\ \frac{\partial g}{\partial x}=\frac{\partial }{\partial x}\left[\frac{\ln \left(y^2+1\right)}{x}\right] \\ \frac{\partial g}{\partial x}=\frac{\partial }{\partial x}\left[\ln \left(y^2+1\right)·x^{-1}\right] \\ \frac{\partial g}{\partial x}=\ln \left(y^2+1\right)\frac{\partial }{\partial x}\left[x^{-1}\right] \\ \frac{\partial g}{\partial x}=\ln \left(y^2+1\right)\left(-1\right)\left(x^{-2}\right) \\ \frac{\partial g}{\partial x}=-\frac{\ln \left(y^2+1\right)}{x^2} \\ g\left(x,y\right)=\frac{\ln \left(y^2+1\right)}{x} \\ \frac{\partial g}{\partial y}=\frac{\partial }{\partial y}\left[\frac{\ln \left(y^2+1\right)}{x}\right] \\ \frac{\partial g}{\partial y}=\frac{1}{x}·\frac{\partial }{\partial y}\left[\ln \left(y^2+1\right)\right] \\ \frac{\partial g}{\partial y}=\frac{1}{x}·\frac{1}{\left(y^2+1\right)}\frac{\partial }{\partial y}\left(y^2+1\right) \\ \frac{\partial g}{\partial y}=\frac{1}{x}·\frac{1}{\left(y^2+1\right)}\left(2y\right) \\ \frac{\partial g}{\partial y}=\frac{2y}{x\left(y^2+1\right)} \\ \text{Hence}, \frac{\partial g}{\partial x}=-\frac{\ln \left(y^2+1\right)}{x^2} \text{and }\frac{\partial g}{\partial y}=\frac{2y}{x\left(y^2+1\right)} \end{gathered}$$