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.

$g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right)鈰�鈰�\left(1\right)$

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

$鈬�\frac{鈭�g}{鈭�x}=\frac{y^2}{1+x^2{y}^4}$

Again partially differentiate both sides with respect to$x$

$鈬�\frac{{鈭�}^{2}g}{鈭�x^2}=-\frac{2x{y}^6}{{\left(1+x^2{y}^4\right)}^2}$

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

$鈬�\frac{鈭�g}{鈭�y}=\frac{2xy}{1+x^2{y}^4}$

Again partially differentiate both sides with respect to $y$

$鈬�\frac{{鈭�}^{2}g}{鈭�y^2}=\frac{2x-6x^3{y}^4}{{\left(1+x^2{y}^4\right)}^2}$

$g\left(x,y\right)={\tan }^{-1}\left(x{y}^2\right)$

$鈬�\frac{{鈭�}^{2}g}{鈭�x鈭�y}=\frac{鈭�}{鈭�x}\left(\frac{2xy}{1+x^2{y}^4}\right)$

role="math" localid="1650463675155" $鈬�\frac{{鈭�}^{2}g}{鈭�x鈭�y}=\frac{2y\left(1-x^2{y}^4\right)}{{\left(1+x^2{y}^4\right)}^2}$

Also

$鈬�\frac{{鈭�}^{2}g}{鈭�y鈭�x}=\frac{鈭�}{鈭�y}\left(\frac{y^2}{1+x^2{y}^4}\right)$

$鈬�\frac{{鈭�}^{2}g}{鈭�y鈭�x}=\frac{2y\left(1-x^2{y}^4\right)}{{\left(1+x^2{y}^4\right)}^2}$

Now as we observe

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