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.

$f\left(x,y\right)=e^x\sin \left(xy\right)⋯⋯\left(1\right)$

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

$⇒\frac{∂f}{∂x}=y{e}^{x}cos\left(xy\right)+e^x\sin \left(xy\right)$

Again partially differentiate both sides with respect to$x$

$⇒\frac{{∂}^{2}f}{∂x^2}=2y{e}^x\cos \left(xy\right)-y^2{e}^x\sin \left(xy\right)+e^x\sin \left(xy\right)$

Partially differentiate equation role="math" localid="1650383152302" $\left(1\right)$both sides with respect to$y$

$⇒\frac{∂f}{∂y}=x{e}^x\cos \left(xy\right)$

Again partially differentiate both sides with respect to $y$

$⇒\frac{{∂}^{2}f}{∂y^2}=-x^2{e}^x\sin \left(xy\right)$

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

$⇒\frac{{∂}^{2}f}{∂x∂y}=\frac{∂}{∂x}\left(x{e}^x\cos \left(xy\right)\right)$

$⇒\frac{{∂}^{2}f}{∂x∂y}=e^x\cos \left(xy\right)+x{e}^x\cos \left(xy\right)-xy{e}^x\sin \left(xy\right)$

Also

$⇒\frac{{∂}^{2}f}{∂y∂x}=\frac{∂}{∂y}\left(y{e}^x\cos \left(xy\right)+e^x\sin \left(xy\right)\right)$

$⇒\frac{{∂}^{2}f}{∂y∂x}=e^x\cos \left(xy\right)-yx{e}^x\sin \left(xy\right)+x{e}^x\cos \left(xy\right)$

Now as we observe

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