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

tabita57i 2021-10-19 Answered
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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Expert Answer

ensojadasH
Answered 2021-10-20 Author has 16248 answers
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
\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{d}}}{{{\left.{d}{y}\right.}}}}{\left({\frac{{{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{x}}}{{{x}^{{2}}+{y}^{{2}}}}}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={x}{\frac{{{\left({x}^{{2}}+{y}^{{2}}\right)}\cdot{1}-{\left({2}{y}+{0}\right)}{y}}}{{{\left({x}^{{2}}+{y}^{{2}}\right)}^{{2}}}}}\)
\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}={x}{\frac{{{x}^{{2}}-{y}^{{2}}}}{{{\left({x}^{{2}}+{y}^{{2}}\right)}^{{2}}}}}\)
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Relevant Questions

asked 2021-10-09
Use implicit differentiation to find \(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}}\) and \(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{y}\right.}}}}\)
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and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.
Let \(\displaystyle{M}={x}{y}^{{{2}}}+{x}\ \text{and}\ {N}={y}{x}^{{{2}}}+{y}\)
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the general solution then is
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Is this solution the same I would get if I had taken the Separate Equations route?
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