`arctan(xy) = arcsin(x + y) , (0,0)` Use implicit differentiation to find an equation of the tangent line at the given point

Saturday, August 14, 2010

$\displaystyle{\arctan{{\left({x}{y}\right)}}}={\arcsin{{\left({x}+{y}\right)}}},{\left({0},{0}\right)}$ Use implicit differentiation to find an equation of the tangent line at the given point

$\displaystyle{\arctan{{\left({x}{y}\right)}}}={\arcsin{{\left({x}+{y}\right)}}}$

First, take the derivative of both sides of the equation using implicit differentiation.

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left[{\arctan{{\left({x}{y}\right)}}}\right]}=\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left({x}+{y}\right)}}}\right]}$

Take note that the derivative formula of arctangent is

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\arctan{{\left({u}\right)}}}\right]}=\frac{{1}}{{{1}+{{u}}^{{2}}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

And the derivative formula of arcsine is

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left({u}\right)}}}\right]}=\frac{{1}}{\sqrt{{{1}-{{u}}^{{2}}}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying these two formulas, the equation becomes

$\displaystyle\frac{{1}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({x}{y}\right)}=\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({x}+{y}\right)}$

To take the derivative of xy, apply the product rule.

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({u}\cdot{v}\right)}={u}\cdot\frac{{{d}{v}}}{{\left.{d}{x}}}+{v}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying this formula, the equation becomes

$\displaystyle\frac{{1}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}\cdot{\left({x}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({y}\right)}+{y}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}\right)}=\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}\cdot{\left(\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}+\frac{{d}}{{\left.{d}{x}}}{\left({y}\right)}\right)}$

$\displaystyle\frac{{1}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}\cdot{\left({x}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}+{y}\cdot{1}\right)}=\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}\cdot{\left({1}+\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}\right)}$

Then, isolate $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}$ .

$\displaystyle\frac{{x}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}+\frac{{y}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}=\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}+\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}$

$\displaystyle\frac{{x}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}-\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}-\frac{{y}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}$

$\displaystyle{\left(\frac{{x}}{{{x}+{{\left({x}{y}\right)}}^{{2}}}}-\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}\right)}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}-\frac{{y}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}-\frac{{y}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}}}{{\frac{{x}}{{{1}+{{\left({x}{y}\right)}}^{{2}}}}-\frac{{1}}{\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}}}\cdot\frac{{\frac{{{\left({1}+{{\left({x}{y}\right)}}^{{2}}\right)}\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}}{{1}}}}{{\frac{{{\left({1}+{{\left({x}{y}\right)}}^{{2}}\right)}\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}}{{1}}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{{\left({1}+{{\left({x}{y}\right)}}^{{2}}\right)}-{y}\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}}{{{x}\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}-{\left({1}+{{\left({x}{y}\right)}}^{{2}}\right)}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{{1}+{{\left({x}{y}\right)}}^{{2}}-{y}\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}}{{{x}\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}-{1}-{{\left({x}{y}\right)}}^{{2}}}}$

Then, plug-in the given point to get the slope of the curve at that point. The given point is (0,0).

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{{1}+{{\left({0}\cdot{0}\right)}}^{{2}}-{0}\cdot\sqrt{{{1}-{{\left({x}+{y}\right)}}^{{2}}}}}}{{{0}\cdot\sqrt{{{1}-{{\left({0}+{0}\right)}}^{{2}}}}-{1}-{{\left({0}\cdot{0}\right)}}^{{2}}}}=\frac{{{1}+{0}+{0}}}{{{0}-{1}-{0}}}=\frac{{1}}{{-{1}}}=-{1}$

Take note that the slope of a curve at point (x,y) is equal to the slope of the line tangent to that point. So the slope of the tangent line is

$\displaystyle{m}=\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=-{1}$

Then, apply the point-slope form to get the equation of the line.

$\displaystyle{y}-{y}_{{1}}={m}{\left({x}-{x}_{{1}}\right)}$

Plugging in the values, it becomes

$\displaystyle{y}-{0}=-{1}{\left({x}-{0}\right)}$

$\displaystyle{y}=-{1}{\left({x}\right)}$

$\displaystyle{y}=-{x}$

Therefore, the equation of the tangent line is $\displaystyle{y}=-{x}$ .

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Saturday, August 14, 2010

$\displaystyle{\arctan{{\left({x}{y}\right)}}}={\arcsin{{\left({x}+{y}\right)}}},{\left({0},{0}\right)}$ Use implicit differentiation to find an equation of the tangent line at the given point

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Saturday, August 14, 2010

Use implicit differentiation to find an equation of the tangent line at the given point

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\arctan{{\left({x}{y}\right)}}}={\arcsin{{\left({x}+{y}\right)}}},{\left({0},{0}\right)}$ Use implicit differentiation to find an equation of the tangent line at the given point

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?