Q 46. - Calculus

Question

In Exercises $43-50$, compute all of the second-order partial derivatives for the functions $g\left(x,y\right)=\frac{\ln \left(y^2+1\right)}{x}$ and show that the mixed partial derivatives are equal. 

Step-by-Step Solution

Answer

The second order partial derivatives for the function are

$$\begin{gathered} \frac{{\partial }^{2}g}{\partial x^2}=\frac{2\ln \left(y^2+1\right)}{x^3} \\ \frac{{\partial }^{2}g}{\partial y^2}=\frac{4\left(1-y^2\right)}{x{\left(y^2+1\right)}^2} \end{gathered}$$

Step 1. Definition

Clairaut's theorem on equality of mixed partials states that under assumption of continuity of both the second-order mixed partials of a function of two variables, the two mixed partials are equal. 

Step 2. Finding second order partial derivative

$g\left(x,y\right)=\frac{\ln \left(y^2+1\right)}{x} \cdots \cdots \left(1\right)$
Partially differentiate equation $\left(1\right)$ both sides with respect to $x$

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

Again partially differentiate both sides with respect to $x$

$\Rightarrow \frac{{\partial }^{2}g}{\partial x^2}=\frac{2\ln \left(y^2+1\right)}{x^3}$

Partially differentiate equation $\left(1\right)$ both sides with respect to $y$

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

Again partially differentiate both sides with respect to $y$

$\Rightarrow \frac{{\partial }^{2}g}{\partial y^2}=\frac{4\left(1-y^2\right)}{x{\left(y^2+1\right)}^2}$

Step 3. Finding mixed order partial derivative

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

$$\begin{gathered} \Rightarrow \frac{{\partial }^{2}g}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{2y}{x\left(y^2+1\right)}\right) \\ \Rightarrow \frac{{\partial }^{2}g}{\partial x\partial y}=-\frac{2y}{x^2\left(y^2+1\right)} \end{gathered}$$

Also

$$\begin{gathered} \Rightarrow \frac{{\partial }^{2}g}{\partial y\partial x}=\frac{\partial }{\partial y}\left(-\frac{\ln \left(y^2+1\right)}{x^2}\right) \\ \Rightarrow \frac{{\partial }^{2}g}{\partial y\partial x}=-\frac{2y}{x^2\left(y^2+1\right)} \end{gathered}$$

Now as we observe

$\frac{{\partial }^{2}g}{\partial x\partial y}=\frac{{\partial }^{2}g}{\partial y\partial x}$ [Hence proved]