Find \frac{dz}{dx} and \frac{dz}{dy}. z=\frac{xy}{x^2+y^2}

Find $$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}$$ and $$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}$$.
$$\displaystyle{z}={\frac{{{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}}$$

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We are given that,
$$\displaystyle{z}={\frac{{{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}}$$
differentiating z w.r.t x assuming y as constant
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}}\right)}$$ Using Quotient Rule and taking y out
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}={y}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{x}}}{{{x}^{{2}}+{y}^{{2}}}}}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}={y}{\frac{{{\left({x}^{{2}}+{y}^{{2}}\right)}\cdot{1}-{\left({2}{x}+{0}\right)}{x}}}{{{\left({x}^{{2}}+{y}^{{2}}\right)}^{{2}}}}}$$
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}={y}{\frac{{{y}^{{2}}-{x}^{{2}}}}{{{\left({x}^{{2}}+{y}^{{2}}\right)}^{{2}}}}}$$
differentiating z w.r.t x assuming y as constant
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{d}}}{{{\left.{d}{y}\right.}}}}{\left({\frac{{{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{x}}}{{{x}^{{2}}+{y}^{{2}}}}}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={x}{\frac{{{\left({x}^{{2}}+{y}^{{2}}\right)}\cdot{1}-{\left({2}{y}+{0}\right)}{y}}}{{{\left({x}^{{2}}+{y}^{{2}}\right)}^{{2}}}}}$$
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={x}{\frac{{{x}^{{2}}-{y}^{{2}}}}{{{\left({x}^{{2}}+{y}^{{2}}\right)}^{{2}}}}}$$