Q 45. - Calculus

Question

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

Step-by-Step Solution

Answer

The second order partial derivatives for the function are

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

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

$f\left(x,y\right)=x^y \cdots \cdots \left(1\right)$

Taking log on both sides

$\Rightarrow \log \left[f\left(x,y\right)\right]=y\log x \cdots \cdots \left(2\right)$
Partially differentiate equation $\left(2\right)$ both sides with respect to $x$

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

Again partially differentiate both sides with respect to $x$

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

Partially differentiate equation $\left(2\right)$ both sides with respect to $y$

$$\begin{gathered} \Rightarrow \frac{1}{f\left(x,y\right)}\frac{\partial f}{\partial y}=\log x \\ \Rightarrow \frac{\partial f}{\partial y}=x^y\log x \end{gathered}$$

Again taking log on both sides

$\Rightarrow \log \left(\frac{\partial f}{\partial y}\right)=y\log x+\log \left(\log x\right)$

Again partially differentiate both sides with respect to $y$

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

Step 3. Finding mixed order partial derivative

$f\left(x,y\right)=x^y$

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

Also

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

Now as we observe

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