91影视

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.

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

Taking log on both sides

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

Again partially differentiate both sides with respect to$x$

$鈬�\frac{{鈭�}^{2}f}{鈭�x^2}=y{x}^{y-2}\left(y-1\right)$

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

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

Again taking log on both sides

$鈬�\log \left(\frac{鈭�f}{鈭�y}\right)=y\log x+\log \left(\log x\right)$

Again partially differentiate both sides with respect to $y$

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

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

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

Also

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

Now as we observe

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