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

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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May Dunn
$$\displaystyle{{\tan}^{{-{1}}}{\left({x}^{{2}}{y}\right)}}={x}+{x}{y}^{{2}}$$
$$\displaystyle{{\tan}^{{-{1}}}{\left({x}^{{2}}{y}\right)}}-{x}-{x}{y}^{{2}}={0}$$
$$\displaystyle{F}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left({x}^{{2}}{y}\right)}}-{x}-{x}{y}^{{2}}={0}$$
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: $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{\left({1}+{y}^{{2}}\right)}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}-{2}{x}{y}}}{{{x}^{{2}}-{2}{x}{y}{\left({1}+{x}^{{4}}{y}^{{2}}\right)}}}}$$