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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 2.6 Implicit Differentiation

Subsection 2.6.1 Before Class

https://mymedia.ou.edu/media/2.6-1/1_iw3gx2a8
Figure 19. Pre-Class Video 1

Subsubsection 2.6.1.1 The Idea

Example 2.6.1.
Consider the equation \(x^2 + y^2 = 1\text{.}\)
  1. \(x^2 + y^2 = 1\) is called an implicit equation. Why is this considered implicit and not explicit?
  2. Write \(x^2 + y^2 = 1\) in an explicit form.
Solution.
  1. \(y\) can’t be written in terms of a single equation, i.e. it’s not a function of \(x\)
  2. \(\displaystyle y = \pm \sqrt{1-x^2}\)
Example 2.6.2.
Consider the circle of radius 4, centered at the origin, given by the implicit equation \(x^2 + y^2 = 16\text{.}\)
  1. Rewrite the equation using that \(y = y(x)\) (that is, \(y\) is a function of \(x\))
  2. Why would we need to use the chain rule in order to take this derivative?
  3. Take the derivative of both sides, using \(\dfrac{dy}{dx}\) for the derivative of \(y(x)\)
  4. Find \(\dfrac{dy}{dx}\text{.}\)
Solution.
  1. \(\displaystyle x^2 + (y(x))^2 = 4\)
  2. There’s function composition
  3. \(\displaystyle 2x+2(y(x))\cdot \dfrac{dy}{dx} = 0\)
  4. \(\displaystyle \dfrac{dy}{dx} = -\dfrac{x}{y(x)} = - \dfrac{x}{y}\)
Example 2.6.3.
Use implicit differentiation to find \(\dfrac{dy}{dx}\) for the implicit equation \(x^3 + y^3 = x\)
Solution.
\(\dfrac{dy}{dx} = \dfrac{1-3x^2}{3(y(x))^2} = \dfrac{1-3x^2}{3y^2}\)

Subsection 2.6.2 Pre-Class Activities

Example 2.6.4.

How can you tell if you are looking at an implicit equation or an explicit equation?
Solution.
Answers vary

Example 2.6.5.

  1. Find \(\dfrac{dy}{dx}\) for the expression \(\sqrt{x} + \sqrt{y} = 1\text{,}\) and find the slope of the tangent line at the point \((0,1)\)
  2. Use your answer in (a) to write an equation for the tangent line to the curve at the point \((0,1)\text{.}\)
Solution.
  1. \(\dfrac{dy}{dx} = -\dfrac{\sqrt{y}}{\sqrt{x}}\text{,}\) so \(\dfrac{dy}{dx}\bigg\rvert_{x=2,y=2} = -1\)
  2. \(\displaystyle y = -x+4\)

Subsection 2.6.3 In Class

Example 2.6.6.

  1. Find the derivative of \(x^2+2y^2=9\) at the point \((1,2)\)
  2. Find the equation of the tangent line to the curve at this point.
Solution.
  1. \(\dfrac{dy}{dx} = -\dfrac{x}{2y}\text{,}\) so \(\dfrac{dy}{dx}\bigg\rvert_{x=1,y=2} = -\dfrac{1}{4}\)
  2. \(\displaystyle y =-\dfrac{1}{4}x + \dfrac{9}{4}\)

Example 2.6.7.

Find all points where the tangent line to the curve \(x^4 + y^4 = 16\) is horizontal.
Solution.
\((0,\pm 2)\)

Example 2.6.8.

Find \(y'\text{,}\) if \(\sin (x-y) = y^2\cos x\)
Solution.
\(y' = \dfrac{\cos(x-y)+y^2\sin x}{2y\cos x-\cos(x-y)}\)

Example 2.6.9.

Find \(y'\text{,}\) if \(\dfrac{x^2}{x+y} = y^2 + 1\)
Solution.
\(y' = \dfrac{2x^2+2xy-x^2}{2y(x+y)^2+x^2}\)

Example 2.6.10.

Find \(\dfrac{dy}{dx}\) if \(\sqrt{xy} = 1 + x^2y\)
Solution.
\(y' = \dfrac{\dfrac{1}{2}(xy)y^{-1/2}-2xy}{x^2-\dfrac{1}{2}(xy)^{-1/2}}\)

Example 2.6.11.

Find the equation of the tangent line to the curve \(y\sin 2x = x\cos 2y\) at the point \(\lrpar{\dfrac{\pi}{2},\dfrac{\pi}{4}}\text{.}\)
Solution.
\(y = \dfrac{1}{2}x\)

Example 2.6.12.

Find \(\dfrac{dy}{dx}\) for the equation \(xy = \sqrt{x^2+y^2}\)
Solution.
\(\dfrac{dy}{dx} = \dfrac{x(x^2+y^2)^{-1/2}-y}{x-y(x^2+y^2)^{-1/2}}\)

Example 2.6.13.

Find \(y''\text{,}\) if \(x^2 + 4y^2 = 4\)
Solution.
\(y'' = \dfrac{-4y-\dfrac{x^2}{y}}{16y^2}\)

Example 2.6.14.

Find \(y''\) if \(\sin y + \cos x = 1\)
Solution.
\(y''= \dfrac{\cos x\cos y + \sin^2 x\tan y}{\cos^2y}\)

Subsection 2.6.4 After Class Activities

Example 2.6.15.

Find \(\dfrac{dy}{dx}\) for the equation \(x^4 + x^2y^2 + y^3 = 5\)
Solution.
\(\dfrac{dy}{dx} = \dfrac{-4x-2xy^2}{2x^2y+3y^2}\)

Example 2.6.16.

Find \(\dfrac{dy}{dx}\) for the equation \(\sqrt{x+y} = x^3 + y^3\)
Solution.
\(\dfrac{dy}{dx} = \dfrac{\dfrac{1}{2}(x+y)^{-1/2}-3x^2}{3y^2-\dfrac{1}{2}(x+y)^{-1/2}}\)

Example 2.6.17.

If \(f(x) + x^2[f(x)]^3 = 10\) and \(f(1) = 2\text{,}\) find \(f'(1)\text{.}\)
Solution.
\(f'(1) = -\dfrac{16}{13}\)

Example 2.6.18.

If \(x^2 + xy + y^3 = 1\text{,}\) find the value of \(y''\) at input \(x = 1\text{.}\)
Solution.
\(y''\rvert_{x=1,y=0} = -2\)

Example 2.6.19.

Show that the tangent line to the ellipse
\begin{equation*} \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \end{equation*}
at the point \((x_0,y_0)\) can be written as
\begin{equation*} \dfrac{xx_0}{a^2} + \dfrac{yy_0}{b^2} = 1 \end{equation*}
Solution.
First note that by clearing fractions we get \(b^2x^2+a^2y^2 = a^2b^2\text{.}\) Taking the implicit derivative, we see that \(2b^2x+2a^2yy' = 0\) so that \(y' = -\dfrac{b^2x}{a^2y}\)
Now, the equation of the tangent line can be written as \(y-y_0 = -\dfrac{b^2}{a^2}\lrpar{\dfrac{x_0}{y_0}}(x-x_0)\text{.}\) Simplifying by clearing fractions gives \(a^2yy_0-a^2y_0^2=-b^2xx_0 + b^2x_0^2\text{.}\) Rewriting as \(a^2yy_0 + b^2xx_0 = b^2x_0^2+a^2y_0^2\) and dividing by \(a^2b^2\) gives
\begin{equation*} \dfrac{xx_0}{a^2} + \dfrac{yy_0}{b^2} = \dfrac{x_0^2}{a_0^2} + \dfrac{y_0^2}{b_0^2} \end{equation*}
Using the relationship \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\text{,}\) we conclude that \(\dfrac{xx_0}{a^2} + \dfrac{yy_0}{b^2} = 1\)
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