2.8: Implicit Differentiation
- Page ID
- 204100
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)2.8 Implicit Differentiation
- Apply implicit differentiation to find the derivative of an implicit equation.
- Use implicit differentiation to calculate the derivative of an implicit function at a given point and apply it to determine the equation of the tangent line at a given point.
Usually, we define \( y \) as an explicit function of \( x \). For example: \(y = \sqrt{4 - x^2}\)
However, the equation
\[
x^2 + y^2 = 4
\]
defines \( y \) implicitly as a function of \( x \).
The graph of the equation can be broken into two pieces when we solve for \( y \):
\[
y = \pm \sqrt{4 - x^2}
\]
What about the derivative of a function given implicitly? First, let's analyze the following example:
Suppose that the explicit formula of \( y = f(x) \) is unknown and that \( f \) is differentiable.
Calculate:
\[
\frac{d}{dx}\left[ x + f(x)^2 \right] = \frac{d}{dx}\left[ x + y^2 \right]
\]
We differentiate both sides with respect to \( x \):
\[
\frac{d}{dx}(x) + \frac{d}{dx}(y^2) = 1 + 2y \cdot \frac{dy}{dx}
\]
So the result is:
\[
\frac{d}{dx}\left[ x + y^2 \right] = 1 + 2y \cdot \frac{dy}{dx}
\]
Problem-Solving Strategy:
Suppose that we define a function \( y \) implicitly in terms of \( x \). To calculate the derivative of \( y \) using implicit differentiation, follow these steps:
- Take the derivative of both sides of the equation with respect to \( x \). For example: \[\frac{d}{dx}(\cos x) = -\sin x \quad \text{but} \quad \frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx}\]
- Rewrite the resulting equation so that all terms involving \( \frac{dy}{dx} \) are on the left-hand side, and all other terms are on the right-hand side.
- Factor out \( \frac{dy}{dx} \) on the left-hand side.
- Solve the equation for \( \frac{dy}{dx} \).
Compute the derivative \( \dfrac{dy}{dx} \) of the following equations (assume that \( y \) is a function of \( x \)):
- \( x^2 + y^2 = 4 \)
- \( x^3 + y^2 - 3xy = 3 \)
- \( x^3 \sin y = x^4 \)
- \( \arctan(x^3y)=xy^3\)
- \( \sqrt{x+y}=1+x^2y^2\)
Find the equations of the tangent lines to the curve \( x^2 + y^2 = 4\) at the following points:
(a) \( (0, 2) \)
(b) \( (-2, 0) \)
(c) \( (1, \sqrt{3}) \)
Find the equation of the tangent line to the curve \(\frac{x^2}{9}+\frac{y^2}{36}=1 \) at the point \( (-2, -2\sqrt{5}) \).
Find the equation of the tangent line to the curve \( x^3 + y^2 - 3xy = 3\) at the point \( (-1, 1) \).