Skip to main content

Functions and single-variable calculus A course resource for Math 218

Ellen Goldstein, Aaron Greicius

Section 1.23 Implicit differentiation

We begin this section with some simple observations about curves \(\mathcal{C}\) in \(\R^2\text{.}\) Currently our main way of defining such a curve is as the graph of an equation. For example, let \(\mathcal{C}\) be the unit circle centered at the origin is defined as the graph of the equation
\begin{equation*} x^2+y^2=1\text{.} \end{equation*}
That is, \(\mathcal{C}\) is the set of all points \(P=(x,y)\) satisfying \(x^2+y^2=1\text{.}\) We call this an implicit description of the circle \(\mathcal{C}\) as its points are not given to us explicitly via some formula; rather, to produce actual points on \(\mathcal{C}\) it is up to us to find solutions to the equation (e.g., \(P=(-1,0)\text{,}\) \(Q=(\sqrt{2}/2, \sqrt{2}/2)\)).
🔗
Graph of unit circle
Figure 1.23.1. Graph of \(x^2+y^2=1\)
🔗
A similar way of defining a curve in \(\R^2\) is as the graph of a function. In other words, given a function \(f\colon D\rightarrow \R\text{,}\) we define its graph to be the curve \(\mathcal{C}\) consisting of all points of the form \(P=(x,f(x))\text{,}\) where \(x\in D\text{.}\) Note that this description gives an explicit formula for producing points on the curve: namely, given any \(a\) in the domain, the point \(P=(a,f(a))\) lies on the graph.
🔗
Graph of function x squared
Figure 1.23.2. Graph of the function \(f(x)=x^2\)
🔗
The graph of a function can also be understood as the graph of an equation: indeed, the set of points of the form \(P=(x,f(x))\) are precisely the the set of solutions \((x,y)\) to the equation
\begin{equation*} y=f(x)\text{.} \end{equation*}
In other words, all curves that are graphs of functions can also be thought of as graphs of equations. However, not all graphs of equations are graphs of functions! Indeed, our circle above is an example of a curve that is a graph of an equation, but not the graph of a function. How do I know? It fails the vertical line test!
🔗
Life is somewhat easier for us when a curve does happen to be given as the graph of a function. For example, in this scenario, we have a systematic way of determining what the tangent line to a given point is. That technique is not in general available to us for the circle \(x^2+y^2=1\text{,}\) as that curve is not the graph of any function \(f\text{.}\) So how can we answer questions about tangent lines to that curve. The next example illustrates a technique that we call implicit differentiation. We give a formal description of the technique in Procedure 1.23.4.
🔗

Example 1.23.3. Implicit differentiation: circle.

Use implicit differentiation to compute an equation of the tangent line to the circle \(\mathcal{C}\colon x^2+y^2=1\) at the point \(P=\left(-\frac{1}{3}, \frac{2\sqrt{2}}{3}\right)\text{.}\)
🔗
Solution.
The segment of \(\mathcal{C}\) consisting of points lying very close to \(P\) does pass the vertical line test, and thus we can think of the \(y\)-coordinate value of points on this segment as being given by a function \(y=y(x)\) of the \(x\)-coordinate. The slope of the tangent line to \(\mathcal{C}\) at \(P\) is then given by \(y'(-1/3)\text{.}\) In order to deduce information about \(y'(x)\) we take the derivative of both sides of the defining equation of \(\mathcal{C}\text{,}\) making sure to treat \(y\) not as an independent variable, but rather a function of \(x\text{:}\)
\begin{align*} x^2+(y(x))^2=1 \amp \implies \frac{d}{dx}(x^2+(y(x))^2)=\frac{d}{dx}(1) \\ \amp \implies 2x+2 y(x) y'(x)=0 \amp \text{(chain rule)}\\ \amp \implies y'(x)=-\frac{x}{y(x)}\text{.} \end{align*}
Evaluating the last equation at \(x=-1/3\text{,}\) we conclude that
\begin{align*} y'(-1/3) \amp = -\frac{-1/3}{y(-1/3)}\\ \amp = -\frac{-1/3}{2\sqrt{2}/3}\\ \amp = \frac{1}{2\sqrt{2}}\\ \amp =\frac{\sqrt{2}}{4}\text{.} \end{align*}
Note that \(y(-1/3)=2\sqrt{2}/3\) since by definition \(y(-1/3)\) is the \(y\)-coordinate of the point \(P=(-1/3, 2\sqrt{2}/3)\text{.}\)
🔗
We conclude that the tangent line to \(\mathcal{C}\) at \(P\) has slope \(y'(-1/3)=\sqrt{2}/4\text{.}\) Thus
\begin{equation*} y-\frac{2\sqrt{2}}{3}=\frac{\sqrt{2}}{4}(x+1/3)\text{.} \end{equation*}
Below you find a graph of \(\mathcal{C}\) and the tangent line to \(P\text{.}\)
🔗
Graph of unit circle
🔗
🔗
We formalize our reasoning in Example 1.23.3 with a procedure.
🔗

Example 1.23.5. Tschirnhausen cubic.

The curve \(\mathcal{C}\) defined by the equation \(x^2=y^3+3y^2\) is called Tschirnhausen’s cubic. (See Figure 1.23.6.)
  1. Verify that \(P=(2,1)\) is a point on \(\mathcal{C}\text{.}\)
    🔗
    🔗
  2. Use implicit differentiation to find an equation of the tangent line to \(\mathcal{C}\) at the point \(P=(2,1)\text{.}\)
    🔗
    🔗
  3. Assuming \(y=y(x)\) is can be understood as a function of \(x\) for points on \(\mathcal{C}\) near \((2,1)\text{,}\) compute \(y''(2)\text{.}\)
    🔗
    🔗
🔗
Solution.
  1. To verify that \(P=(2,1)\) lies on \(\mathcal{C}\text{,}\) we simply evaluate the left- (LHS)and right-hand sides (RHS) of the defining equation and verify that they are equal:
    \begin{align*} \text{LHS}\amp = 4 \\ \text{RHS} \amp = 1^3+3\cdot 1^2=4\text{.} \end{align*}
    🔗
    🔗
  2. Following Procedure 1.23.4 we write \(y=y(x)\) for each instance of \(y\) in the defining equation of \(\mathcal{C}\) and then take the derivative of both sides of the equation with respect to \(x\text{:}\)
    \begin{align*} x^2=(y(x))^3+3(y(x))^2 \amp \implies \frac{d}{dx}(x^2)=\frac{d}{dx}((y(x))^3+3(y(x))^2) \\ \amp \implies 2x=3y(x)^2y'(x)+6y(x)y'(x)\\ \amp \implies y'(x)=\frac{2x}{3y(x)^2+6y(x)}\text{.} \end{align*}
    The slope of the tangent line to \(\mathcal{C}\) at \(P\) is thus given by
    \begin{align*} y'(2) \amp =\frac{2\cdot 2}{3(y(2)^2+6y(2))}\\ \amp = \frac{4}{3\cdot 1^2+6\cdot 1}\\ \amp = \frac{4}{9}\text{.} \end{align*}
    The equation of the tangent line, using point-slope form for this \(P\text{,}\) is
    \begin{equation*} y-1=\frac{4}{9}(x-2)\text{.} \end{equation*}
    Below you find a graph of \(\mathcal{C}\) and the tangent line to \(\mathcal{C}\) at \(P\text{.}\)
    🔗
    🔗
  3. To deduce information about \(y''(x)\) we take the derivative with respect to \(x\) of both sides of the equation we derived for \(y'(x)\text{:}\)
    \begin{align*} y'(x)=\frac{2x}{3y(x)^2+6y(x)} \amp \implies \frac{d}{dx}(y'(x))=\frac{d}{dx}\left(\frac{2x}{3y(x)^2+6y(x)}\right)\\ \amp \implies y''(x)=\frac{(2x)'(3y(x)^2+6y(x))-(2x)(3y(x)^2+6y(x))'}{(3y(x)^2+6y(x))^2}\\ \amp \implies y''(x)=\frac{2(3y(x)^2+6y(x))-(2x)(6y(x)y'(x)+6y'(x))}{(3y(x)^2+6y(x))^2}\text{.} \end{align*}
    We then compute
    \begin{align*} y''(2) \amp = \frac{2(3y(2)^2+6y(2))-(2\cdot 2)(6y(2)y'(2)+6y'(2))}{(3y(2)^2+6y(2))^2}\\ \amp = \frac{2(3\cdot 1^2+6\cdot 1)-4(6\cdot 1\cdot (4/9)+6\cdot (4/9))}{(3\cdot 1^2+6\cdot 1)^2} \amp (y(2)=1, y'(2)=4/9)\\ \amp = \frac{18-4(48/9)}{81}\\ \amp = \frac{162-192}{729}\\ \amp = -\frac{30}{729}\\ \amp = -\frac{10}{243}\text{.} \end{align*}
    🔗
    🔗
🔗
🔗
🔗
Figure 1.23.6. Tangent line to a point on Tschirnhausen’s cubic
🔗

Example 1.23.7. Ellipse tangents.

Let \(\mathcal{C}\) be the ellipse defined by the equation \(5x^2-6xy+2y^2=1\text{.}\) Find all points \(P\) on \(\mathcal{C}\) where the tangent line to \(\mathcal{C}\) at \(P\) is horizontal. (See Figure 1.23.8.)
🔗
Solution.
Applying Procedure 1.23.4 we have:
\begin{align*} 5x^2-6xy(x)+2(y(x))^2=1 \\ \frac{d}{dx}(5x^2-6xy(x)+2(y(x))^2)\amp =\frac{d}{dx}(1) \amp \\ 10x-6(x\, y(x))'+4y(x)y'(x)\amp =0\\ 10x-6(y+xy'(x))+4yy'(x)=0\text{.} \end{align*}
We look for points \(P=(x,y)\) where \(y'(x)=0\text{,}\) making the tangent line horizontal. To see which points satisfy this, we setting \(y'(x)=0\) in the equation above and see what it says:
\begin{align*} 10x-6(y+0)+0 \amp =0 \\ 10x-6y \amp =0 \\ y \amp =\frac{5}{3}x\text{.} \end{align*}
Thus, if \(P=(x,y)\) is a point on \(\mathcal{C}\) where the tangent line is horizontal, then we must have \(y=5x/3\text{.}\) Substitution this expression for \(y\) back into the defining equation, we have
\begin{align*} 5x^2-6x(5x/3)+2(5x/3)^2\amp =1 \amp \\ \frac{5}{9}x^2\amp =1 \\ x \amp =\pm\frac{3}{\sqrt{5}}\text{.} \end{align*}
The choice \(x=\frac{3}{\sqrt{5}}\) gives rise to the point
\begin{equation*} P_1=\left( \frac{3}{\sqrt{5}}, \frac{5}{\sqrt{5}}\right)\text{.} \end{equation*}
The choice \(x=-\frac{3}{\sqrt{5}}\) gives rise to the point
\begin{equation*} P_2=\left(-\frac{3}{\sqrt{5}}, -\frac{5}{\sqrt{5}}\right)\text{.} \end{equation*}
We conclude that \(P_1\) and \(P_2\) are the points where the tangent line to \(\mathcal{C}\) is horizontal.
🔗
🔗
🔗
A rotated ellipse
Figure 1.23.8. Horizontal tangent lines on ellipse \(5x^2-6xy+2y^2=1\)
🔗

Example 1.23.9. Implicit: radical equation.

Let the curve \(\mathcal{C}\) be defined by the equation \(\sqrt{y^2+9}+xy^2=11x\text{.}\) (See Figure 1.23.10)
  1. Verify that \(P=(-1,4)\) is a point on this curve.
    🔗
    🔗
  2. Find an equation of the tangent line to \(\mathcal{C}\) at \(P=(-1,4)\text{.}\)
    🔗
    🔗
🔗
Solution.
  1. It is easily seen that the point \(P=(0,0)\) satisfies the defining equation of \(\mathcal{C}\text{.}\)
    🔗
    🔗
  2. We apply Procedure 1.23.4:
    \begin{align*} \frac{d}{dx}(\sqrt{y^2+9}+xy^2) \amp =\frac{d}{dx}(11x)\\ \frac{(y^2+9)'}{2\sqrt{y^2+9}}+(x)'y^2+x(y^2)' \amp = 11 \\ \frac{2y\cdot y'(x)}{2\sqrt{y^2+9}}+y^2+2xy\,y'(x) \amp = 11\\ \frac{y\, y'(x)}{\sqrt{y^2+9}}+y^2+2xy\, y'(x) \amp = 11\text{.} \end{align*}
    Evaluating this equation at the point \(P=(-1,4)\text{,}\) we conclude that
    \begin{align*} \frac{4y'(x)}{5}+16-8y'(x) \amp =11\\ 4y'(x)+80 -40y'(x) \amp =55\\ y'(x)\amp =\frac{25}{36} \amp \text{.} \end{align*}
    It follows that the tangent line to \(\mathcal{C}\) at \(P=(-1,4)\) has defining equation \(y-4=-\frac{39}{20}(x+1)\text{.}\)
    🔗
    🔗
🔗
🔗
🔗
Figure 1.23.10. Tangent to curve \(\sqrt{y^2+9}+xy^2=11x\)
🔗
🔗
PrevTopNext