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### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Use implicit differentiation to find an equation of the tangent line to the curve at the given point.22. $${{\rm{x}}^{{\raise0.7ex\hbox{{\rm{2}}} \!\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ + }}{{\rm{y}}^{{\raise0.7ex\hbox{{\rm{2}}}\!\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ = 4, }}\left( {{\rm{ - 3}}\sqrt {\rm{3}}{\rm{,1}}} \right)$$ (astroid)

The equation of the tangent line at $$\left( {{\rm{ - 3}}\sqrt {\rm{3}} {\rm{,1}}} \right)$$is $${\rm{y = }}\frac{{\rm{1}}}{{\sqrt {\rm{3}} }}{\rm{x + 4}}$$

See the step by step solution

## Step$${\rm{1}}$$: Given information

The given equation is $${{\rm{x}}^{{\raise0.7ex\hbox{{\rm{2}}} \!\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ + }}{{\rm{y}}^{{\raise0.7ex\hbox{{\rm{2}}}\!\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ = 4}}$$ with the point $$\left( {{\rm{ -3}}\sqrt {\rm{3}} {\rm{,1}}} \right)$$

## Step$${\rm{2}}$$: Definition of implicit differentiation

Implicit differentiation: This method consists of differentiating both sides of the equation with respect to $${\rm{x}}$$ and then solving the resulting equation for $${\rm{y'}}$$

## Step$${\rm{3}}$$: Differentiate the given equation with respect to $${\rm{x}}$$

The equation is

$${{\rm{x}}^{{\raise0.7ex\hbox{{\rm{2}}} \!\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ + }} {{\rm{y}}^{{\raise0.7ex\hbox{{\rm{2}}}\!\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ = 4,}}$$

Differentiate both sides with respect to $${\rm{x}}$$

$$\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{{\raise0.7ex\hbox{{\rm{2}}} \!\mathord{\left/ {\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ + }}{{\rm{y}}^{{\raise0.7ex\hbox{{\rm{2}}} \!\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}} \right){\rm{ = }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{4}} \right)$$

The Sum rule for differentiation

$$\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right){\rm{ + g}}\left( {\rm{x}} \right)} \right){\rm{ = }}\frac{{{\rm{d}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}}{{{\rm{dx}}}}{\rm{ + }}\frac{{{\rm{d}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)}}{{{\rm{dx}}}}$$

$$\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{{\raise0.7ex\hbox{{\rm{2}}} !\mathord{\left/{\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}} \right){\rm{ + }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left({{{\rm{y}}^{{\raise0.7ex\hbox{{\rm{2}}} \!\mathord{\left/ {\vphantom {{\rm{2}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}} \right){\rm{ = 0}}$$

Chain rule: The chain rule states that the derivative of

$${\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)$$ is equal to $${\rm{f'}}\left({{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{ \times g'}}\left( {\rm{x}} \right)$$

\begin{aligned}\frac{2}{3}{{\rm{x}}^{{\raise0.7ex\hbox{{ - 1}} \!\mathord{\left/{\vphantom {{ 1}{\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ + }}\frac{2}{3}{{\rm{y}}^{{\raise0.7ex\hbox{{ - 1}} \!\mathord{\left/{\vphantom {{ - 1}{\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{y' = 0}}\\{{\rm{x}}^{{\raise0.7ex\hbox{{ -1}} \!\mathord{\left/{\vphantom {{ - 1} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ +}}{{\rm{y}}^{{\raise0.7ex\hbox{{ - 1}} \!\mathord{\left/{\vphantom {{ - 1}{\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{y' = 0}}\end{aligned}

Take $${\rm{y'}}$$ common from the two terms in left hand side

\begin{aligned}{{\rm{y}}^{{\raise0.7ex\hbox{{{\rm{ - 1}}}} \!\mathord{\left/ {\vphantom {{{\rm{ -1}}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}y' &= - {{\rm{x}}^{{\raise0.7ex\hbox{{{\rm{ - 1}}}}\!\mathord{\left/ {\vphantom {{{\rm{ - 1}}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}\\y' &= \frac{{{\rm{ - }}{{\rm{x}}^{{\raise0.7ex\hbox{{{\rm{ - 1}}}} \!\mathord{\left/ {\vphantom {{{\rm{ - 1}}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}}}{{{{\rm{y}}^{{\raise0.7ex\hbox{{{\rm{ - 1}}}} \!\mathord{\left/ {\vphantom {{{\rm{ - 1}}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}}}& = -{\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)^{{\raise0.7ex\hbox{{{\rm{ - 1}}}} \!\mathord{\left/{\vphantom {{{\rm{ - 1}}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}&= - {\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right)^{{\raise0.7ex\hbox{{\rm{1}}} \!\mathord{\left/ {\vphantom {{\rm{1}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}\end{aligned}

## Step$${\rm{4}}$$: Find the slope of the tangent line at $$\left( {{\rm{ - 3}}\sqrt {\rm{3}} {\rm{,1}}} \right)$$

The tangent line of a curve at a given point is a line that just touches the curve at that point and the slope of the tangent line of $${\rm{y = f}}\left( {\rm{x}} \right)$$ at a point $$\left( {{{\rm{x}}_{\rm{0}}}{\rm{,}}{{\rm{y}}_{\rm{0}}}} \right)$$ is $${\left. {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right|_{\left( {{{\rm{x}}_{\rm{0}}}{\rm{,}}{{\rm{y}}_{\rm{0}}}} \right)}}$$

So the slope of the tangent line at $$\left( {{\rm{ - 3}}\sqrt {\rm{3}} {\rm{,1}}} \right)$$ is

$${\left. {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right|_{\left( {{\rm{ - 3}}\sqrt {\rm{3}} {\rm{,1}}} \right)}}{\rm{ = - }}{\left( {\frac{{\rm{1}}}{{{\rm{ - 3}}\sqrt {\rm{3}} }}} \right)^{{\raise0.7ex\hbox{{\rm{1}}} \!\mathord{\left/{\vphantom {{\rm{1}} {\rm{3}}}}\right.}\!\lower0.7ex\hbox{{\rm{3}}}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{3}} }}$$

## Step$${\rm{5}}$$: Find an equation of the tangent line at $$\left( {{\rm{ - 3}}\sqrt {\rm{3}} {\rm{,1}}} \right)$$

The tangent line formula is,

$${\rm{y - }}{{\rm{y}}_{\rm{0}}}{\rm{ = }}{\left. {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right|_{\left( {{{\rm{x}}_{\rm{0}}}{\rm{,}}{{\rm{y}}_{\rm{0}}}} \right)}}\left( {{\rm{x - }}{{\rm{x}}_{\rm{0}}}} \right)$$

An equation of the tangent line at $$\left( {{\rm{ - 3}}\sqrt {\rm{3}} {\rm{,1}}} \right)$$ is

\begin{aligned}y - 1 &= \frac{{\rm{1}}}{{\sqrt {\rm{3}} }}\left( {{\rm{x - }}\left( {{\rm{ - 3}}\sqrt {\rm{3}} } \right)} \right)\\y - 1 &=\frac{{\rm{1}}}{{\sqrt {\rm{3}} }}{\rm{x + 3}}\\y &= \frac{{\rm{1}}}{{\sqrt {\rm{3}} }}{\rm{x + 4}}\end{aligned}

## Step$${\rm{6}}$$: Plot the graph

Therefore, the equation of the tangent line at $$\left( {{\rm{ - 3}}\sqrt {\rm{3}} {\rm{,1}}} \right)$$is $${\rm{y = }}\frac{{\rm{1}}}{{\sqrt {\rm{3}} }}{\rm{x + 4}}$$