91Ó°ÊÓ

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.

$g\left(x,y\right)=x\cos y⋯⋯\left(1\right)$
Partially differentiate equation $\left(1\right)$both sides with respect to$x$

$⇒\frac{∂g}{∂x}=\cos y$

Again partially differentiate both sides with respect to$x$

$⇒\frac{{∂}^{2}g}{∂x^2}=0$

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

role="math" localid="1650472260509" $⇒\frac{∂g}{∂y}=-x\sin y$

Again partially differentiate both sides with respect to $y$

$⇒\frac{{∂}^{2}g}{∂y^2}=-x\cos y$

$g\left(x,y\right)=x\cos y$

$$\begin{gathered} ⇒\frac{{∂}^{2}g}{∂x∂y}=\frac{∂}{∂x}\left(-x\sin y\right) \\ ⇒\frac{{∂}^{2}g}{∂x∂y}=-\sin y \end{gathered}$$

Also

$$\begin{gathered} ⇒\frac{{∂}^{2}g}{∂y∂x}=\frac{∂}{∂y}\left(cosy\right) \\ ⇒\frac{{∂}^{2}g}{∂y∂x}=-\sin y \end{gathered}$$

Now as we observe

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