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Question Number 116319 by bemath last updated on 03/Oct/20

$$\mathrm{If}\mathrm{z}={\mathrm{x}}^2{\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right),\mathrm{find}\frac{{\partial }^2\mathrm{z}}{\partial \mathrm{x}\partial \mathrm{y}}$$$$\mathrm{at}(1,1)$$

Answered by john santu last updated on 03/Oct/20

$$\frac{\partial z}{\partial y}=x^2.\frac{x^2}{x^2+y^2}.\frac{\partial }{\partial y}\left(\frac{y}{x}\right)=\frac{x^3}{x^2+y^2}$$$$\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial }{\partial x}\left(\frac{x^3}{x^2+y^2}\right)$$$$=\frac{3x^2(x^2+y^2)-x^3\left(2x\right)}{{(x^2+y^2)}^2}$$$$=\frac{x^4+3x^2{y}^2}{{(x^2+y^2)}^2}{\mid }_{(1,1)}=\frac{1+3}{{(1+1)}^2}$$$$=1.$$

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