Q 44. - Calculus

Question

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

Step-by-Step Solution

Answer

The second order partial derivatives for the function are

$$\begin{gathered} \frac{{\partial }^{2}g}{\partial x^2}=-\frac{2x{y}^6}{{\left(1+x^2{y}^4\right)}^2} \\ \frac{{\partial }^{2}g}{\partial y^2}=-\frac{2x-6x^3{y}^4}{{\left(1+x^2{y}^4\right)}^2} \end{gathered}$$

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

$g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right) \cdots \cdots \left(1\right)$

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

$\Rightarrow \frac{\partial g}{\partial x}=\frac{y^2}{1+x^2{y}^4}$

Again partially differentiate both sides with respect to $x$

$\Rightarrow \frac{{\partial }^{2}g}{\partial x^2}=-\frac{2x{y}^6}{{\left(1+x^2{y}^4\right)}^2}$

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

$\Rightarrow \frac{\partial g}{\partial y}=\frac{2xy}{1+x^2{y}^4}$

Again partially differentiate both sides with respect to $y$

$\Rightarrow \frac{{\partial }^{2}g}{\partial y^2}=\frac{2x-6x^3{y}^4}{{\left(1+x^2{y}^4\right)}^2}$

Step 3. Finding mixed order partial derivative

$g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right)$

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

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

Also

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

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

Now as we observe

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