## Calculus: Early Transcendentals 8th Edition

$u_{xy}=-2x^{3}ycos(x^{2}y)-2xsin(x^{2}y)$ and $u_{yx}=-2x^{3}ycos(x^{2}y)-2xsin(x^{2}y)$ Hence, $u_{xy}=u_{yx}$
Consider the function $u=cos(x^{2}y)$ Need to prove the conclusion of Clairaut’s Theorem holds, that is, $u_{xy}=u_{yx}$ In order to find this differentiate the function with respect to $x$ keeping $y$ constant. $u_{x}=-2xysin(x^{2}y)$ Differentiate the function with respect to $x$ keeping $y$ constant. $u_{y}=-sin(x^{2}y)x^{2}$ Differentiate $u_{x}$ with respect to $y$ keeping $x$ constant. $u_{xy}=\frac{∂}{∂y}[-2xysin(x^{2}y)]$ (Apply product rule) $=-2xy\frac{∂}{∂y}[sin(x^{2}y)]+sin(x^{2}y)\frac{∂}{∂y}[(-2xy)]$ $=-x^{2}cos(x^{2}y)2xy-2xsin(x^{2}y)$ Thus, $u_{xy}=-2x^{3}ycos(x^{2}y)-2xsin(x^{2}y)$ Differentiate $u_{y}$ with respect to $x$ keeping $y$ constant. $u_{yx}=\frac{∂}{∂x}[-sin(x^{2}y)x^{2}]$ (Apply product rule) $=-2x^{3}ycos(x^{2}y)-2xsin(x^{2}y)$ Therefore, $u_{xy}=-2x^{3}ycos(x^{2}y)-2xsin(x^{2}y)$ and $u_{yx}=-2x^{3}ycos(x^{2}y)-2xsin(x^{2}y)$ Hence, $u_{xy}=u_{yx}$