Найти частное производное второго порядка

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[latex]z=ln(x+\sqrt{x^2+y^2})\\\\z'_{x}= \frac{1}{x+\sqrt{x^2+y^2}}\cdot (1+\frac{2x}{2\cdot \sqrt{x^2+y^2}} )= \frac{1}{x+\sqrt{x^2+y^2}}\cdot \frac{\sqrt{x^2+y^2}+x}{\sqrt{x^2+y^2}}=\\\\=\frac{1}{\sqrt{x^2+y^2}} \; ;\\\\z''_{xx}= \frac{\frac{2x}{2\sqrt{x^2+y^2}}}{x^2+y^2} = \frac{x}{\sqrt{(x^2+y^2)^3}} \; ;\\\\z'_{y}= \frac{1}{x+\sqrt{x^2+y^2}}\cdot \frac{2y}{2\sqrt{x^2+y^2}}= \frac{y}{\sqrt{x^2+y^2}\cdot (x+\sqrt{x^2+y^2})} \; ;[/latex] [latex]z''_{yy}= \frac{1}{(x^2+y^2)\cdot (x+\sqrt{x^2+y^2})^2} \cdot \Big (\sqrt{x^2+y^2}\cdot (x+\sqrt{x^2+y^2})+\\\\+y\cdot \Big (\frac{2y}{2\sqrt{x^2+y^2}}\cdot \sqrt{x^2+y^2}+(x+\sqrt{x^2+y^2})\cdot \frac{2y}{2\sqrt{x^2+y^2}} \Big )\Big )[/latex] [latex]z''_{xy}=- \frac{\frac{2y}{2\sqrt{x^2+y^2}}}{(x^2+y^2)^3} =-\frac{y}{\sqrt{(x^2+y^2)^3}}[/latex]