Use Equation to find \frac{dy}{dx} \tan^{-1}(x^2y)=x+xy^2

gabioskay7 2021-11-15 Answered
Use Equation to find \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{{\tan}^{{-{1}}}{\left({x}^{{2}}{y}\right)}}={x}+{x}{y}^{{2}}\)

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Expert Answer

May Dunn
Answered 2021-11-16 Author has 7135 answers
\(\displaystyle{{\tan}^{{-{1}}}{\left({x}^{{2}}{y}\right)}}={x}+{x}{y}^{{2}}\)
\(\displaystyle{{\tan}^{{-{1}}}{\left({x}^{{2}}{y}\right)}}-{x}-{x}{y}^{{2}}={0}\)
\(\displaystyle{F}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left({x}^{{2}}{y}\right)}}-{x}-{x}{y}^{{2}}={0}\)
Suppose that an equation \(\displaystyle{F}{\left({x},{y}\right)}={0}\) defines y Implicity as a differentiable function of x.
Then
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{F}_{{x}}}}{{{F}_{{y}}}}}\)
\(\displaystyle{F}_{{x}}={\frac{{{1}}}{{{1}+{\left({x}^{{2}}{y}\right)}^{{2}}}}}\times{\left({2}{x}{y}\right)}-{1}-{y}^{{2}}\)
\(\displaystyle{F}_{{x}}={\frac{{{2}{x}{y}}}{{{1}+{x}^{{4}}{y}^{{2}}}}}-{1}-{y}^{{2}}\)
\(\displaystyle{F}_{{x}}={\frac{{{2}{x}{y}-{\left({1}+{y}^{{2}}\right)}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}}}{{{1}+{x}^{{4}}{y}^{{2}}}}}\)
\(\displaystyle{F}_{{y}}={\frac{{{x}^{{2}}}}{{{1}+{\left({x}^{{2}}{y}\right)}^{{2}}}}}-{2}{x}{y}\)
\(\displaystyle{F}_{{y}}={\frac{{{x}^{{2}}-{2}{x}{y}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}}}{{{1}+{x}^{{4}}{y}^{{2}}}}}\)
Therefore,
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{F}_{{x}}}}{{{F}_{{y}}}}}=-{\frac{{{\frac{{{2}{x}{y}-{\left({1}+{y}^{{2}}\right)}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}}}{{{1}+{x}^{{4}}{y}^{{2}}}}}}}{{{\frac{{{x}^{{2}}-{2}{x}{y}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}}}{{{1}+{x}^{{4}}{y}^{{2}}}}}}}}={\frac{{{\left({1}+{y}^{{2}}\right)}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}-{2}{x}{y}}}{{{x}^{{2}}-{2}{x}{y}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}}}}\)
Result: \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{\left({1}+{y}^{{2}}\right)}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}-{2}{x}{y}}}{{{x}^{{2}}-{2}{x}{y}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}}}}\)
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