Страница сетевой поддержки учеников Тимура Юрьевича Альпина

Найти все производные $z\left(x,y\right)$ первого и второго порядков,
если
$$z=\sqrt{x^{2}-y^{2}}\mathrm{tg}\frac{z}{\sqrt{x^{2}-y^{2}}}$$

Сразу дифференцировать это уравнение вредно. Сначала разделим обе части на $\sqrt{x^{2}-y^{2}}$

$$\frac{z}{\sqrt{x^{2}-y^{2}}}=\mathrm{tg}\frac{z}{\sqrt{x^{2}-y^{2}}}$$ и перенесём всё влево: $$\mathrm{tg}\frac{z}{\sqrt{x^{2}-y^{2}}}-\frac{z}{\sqrt{x^{2}-y^{2}}}=0.$$ Представим уравнение в виде $$F\left(u\right)=0,$$ где $$F\left(u\right)=\mathrm{tg}u-u,\qquad u=\frac{z}{\sqrt{x^{2}-y^{2}}}=z\left(x^{2}-y^{2}\right)^{-1/2}.$$ Тогда после дифференцирования можно сократить $F'\left(u\right)$: $$F'\left(u\right)u'_{x}=0\qquad\Longrightarrow\qquad u'_{x}=0.$$ Далее $$u'_{x}=z'_{x}\left(x^{2}-y^{2}\right)^{-1/2}+z\frac{\partial}{\partial x}\left[\left(x^{2}-y^{2}\right)^{-1/2}\right]=z'_{x}\left(x^{2}-y^{2}\right)^{-1/2}-\frac{1}{2}z\left(x^{2}-y^{2}\right)^{-3/2}2x=$$ $$=z'_{x}\left(x^{2}-y^{2}\right)^{-1/2}-xz\left(x^{2}-y^{2}\right)^{-3/2}=0,$$ $$z'_{x}\left(x^{2}-y^{2}\right)^{-1/2}=xz\left(x^{2}-y^{2}\right)^{-3/2},$$ $$z'_{x}=xz\left(x^{2}-y^{2}\right)^{-2/2}=\frac{xz}{x^{2}-y^{2}}.$$ Аналогично получим $z'_{y}$: $$F'\left(u\right)u'_{y}=0\qquad\Longrightarrow\qquad u'_{y}=0;$$ $$u'_{y}=z'_{y}\left(x^{2}-y^{2}\right)^{-1/2}+z\frac{\partial}{\partial y}\left[\left(x^{2}-y^{2}\right)^{-1/2}\right]=z'_{y}\left(x^{2}-y^{2}\right)^{-1/2}-\frac{1}{2}z\left(x^{2}-y^{2}\right)^{-3/2}\left(-2y\right)=$$ $$=z'_{y}\left(x^{2}-y^{2}\right)^{-1/2}+yz\left(x^{2}-y^{2}\right)^{-3/2}=0,$$ $$z'_{y}\left(x^{2}-y^{2}\right)^{-1/2}=-yz\left(x^{2}-y^{2}\right)^{-3/2},$$ $$z'_{y}=-yz\left(x^{2}-y^{2}\right)^{-2/2}=-\frac{yz}{x^{2}-y^{2}}=\frac{yz}{y^{2}-x^{2}}.$$ Теперь – вторые производные (они считаются уже в лоб): $$z''_{xx}=\frac{\partial}{\partial x}z'_{x}=\frac{x}{x^{2}-y^{2}}z'_{x}+\frac{z}{x^{2}-y^{2}}-\frac{xz}{\left(x^{2}-y^{2}\right)^{2}}2x=\frac{x}{x^{2}-y^{2}}\frac{xz}{x^{2}-y^{2}}+\frac{z\left(x^{2}-y^{2}\right)}{\left(x^{2}-y^{2}\right)^{2}}-\frac{2x^{2}z}{\left(x^{2}-y^{2}\right)^{2}}=$$ $$=\frac{x^{2}z}{\left(x^{2}-y^{2}\right)^{2}}+\frac{z\left(x^{2}-y^{2}\right)}{\left(x^{2}-y^{2}\right)^{2}}-\frac{2x^{2}z}{\left(x^{2}-y^{2}\right)^{2}}=-\frac{y^{2}z}{\left(x^{2}-y^{2}\right)^{2}},$$ $$z'_{yy}=\frac{\partial}{\partial y}z'_{y}=\frac{y}{y^{2}-x^{2}}z'_{y}+\frac{z}{y^{2}-x^{2}}-\frac{yz}{\left(y^{2}-x^{2}\right)^{2}}2y=\frac{y}{y^{2}-x^{2}}\frac{yz}{y^{2}-x^{2}}+\frac{z\left(y^{2}-x^{2}\right)}{\left(y^{2}-x^{2}\right)^{2}}-\frac{2y^{2}z}{\left(y^{2}-x^{2}\right)^{2}}=$$ $$=\frac{y^{2}z}{\left(y^{2}-x^{2}\right)^{2}}+\frac{z\left(y^{2}-x^{2}\right)}{\left(y^{2}-x^{2}\right)^{2}}-\frac{2y^{2}z}{\left(y^{2}-x^{2}\right)^{2}}=-\frac{x^{2}z}{\left(y^{2}-x^{2}\right)^{2}},$$ $$z''_{xy}=\frac{\partial}{\partial y}z'_{x}=\frac{\partial}{\partial y}\left(\frac{xz}{x^{2}-y^{2}}\right)=\frac{x}{x^{2}-y^{2}}z'_{y}-\frac{xz}{\left(x^{2}-y^{2}\right)^{2}}\left(-2y\right)=\frac{x}{x^{2}-y^{2}}\frac{yz}{y^{2}-x^{2}}+\frac{2xyz}{\left(x^{2}-y^{2}\right)^{2}}=$$ $$=-\frac{xyz}{\left(x^{2}-y^{2}\right)^{2}}+\frac{2xyz}{\left(x^{2}-y^{2}\right)^{2}}=\frac{xyz}{\left(x^{2}-y^{2}\right)^{2}}.$$