4.6: Implicit differentiation.

In Section 4.4, Applications of the Power Chain Rule, we found slopes of circles and ellipses. There is a procedure for finding these slopes that requires less algebra, but more mathematical sophistication. In each of the equations,

\[\begin{aligned}
x^{2}+y^{2}&=13\\
\frac{x^{2}}{18}+\frac{y^{2}}{8} &=1 \\
\frac{2 x^{2}}{35}+\frac{3 y^{2}}{35} &=1
\end{aligned}\]

we might solve for \(y\) in terms of \(x\), being careful to choose the correct square root to match the point of tangency, and then compute \(y ^{\prime} (x)\).

In \(x ^{2} + y ^{2} = 13\), we may chose \(y(x) = \sqrt{13 − x ^{2}}\) . Note that

\[x^{2}+y^{2}=x^{2}+\left(\sqrt{13-x^{2}}\right)^{2}=x^{2}+13-x^{2}=13\]

so that \(y(x) = \sqrt{13 − x ^{2}}\) is a function that 'satisfies' and is said to be implicitly defined by the equation.

Definition 4.6.1 Implicit Function.

Suppose we are given an equation

\[E(x, y) = 0\]

and point (a, b) for which

\[E(a, b) = 0\]

A function, \(f\), defined on an interval \((a-h, a+h)\) surrounding a and satisfying

\[E(x, f(x))=0 \quad \text { and } \quad f(a)=b\]

is said to be implicitly defined by \(E\). There may be no such function, \(f\), one such function, or many such functions.

Now we assume without solving for \(y(x)\) that there is a function \(y(x)\) for which

\[x^{2}+(y(x))^{2}=13 \quad \text { and } \quad y(2)=3\]

and use the power rule and power chain rule to differentiate the terms in the equation, as follows.

\[\begin{aligned}
x^{2}+(y(x))^{2} &=13 \\
\left[x^{2}\right]^{\prime}+\left[(y(x))^{2}\right]^{\prime} &=[13]^{\prime} \\
2 x+2 y(x) y^{\prime}(x) &=0
\end{aligned}\]

\[\begin{array}\
\text{The } \textbf{power rule} \text{ is used for} & \text{The } \textbf{power chain rule} \text{ is used for}\\
\left[x^{2}\right]^{\prime}=2 x & \left[(y(x))^{2}\right]^{\prime}=2 y(x) y^{\prime}(x)\\
\end{array}\]

We use \(x = 2\) and \(y(2) = 3\) in the last equation to get

\[2 \times 2+2 \times 3 y^{\prime}(2)=0\]

and solve for \(y ^{\prime} (2)\) to get

\[y^{\prime}(2)=-\frac{2}{3}\]

as was found in Example 4.4.1 to be the slope of the tangent to \(x ^{2} + y ^{2} = 13\) at (2,3).

It is important to remember in the above steps the [ ]\(^{\prime}\) means derivative with respect to the independent variable, \(x\). The Leibnitz notation, \(\frac{d}{dx}\) , explicitly shows this and may be easier to use. We repeat this problem with Leibnitz notation.

\[\begin{aligned}
x^{2}+(y(x))^{2} &=13 \\
\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(y(x))^{2} &=\frac{d}{d x}(13) \\
2 x+2 y(x) \frac{d}{d x} y(x) &=0
\end{aligned}\]

\[\begin{array}\
\text{The } \textbf{power rule} \text{ is used for} & \text{The } \textbf{power chain rule} \text{ is used for}\\
\frac{d}{d x}\left(x^{2}\right)=2 x & \frac{d}{d x}(y(x))^{2}=2 y(x) \frac{d}{d x} y(x)\\
\end{array}\]

Example 4.6.1 We consider another example of implicit differentiation. Find the slope of the graph of

\[\sqrt{x}+\sqrt{5-y^{2}}=5 \quad \text { at }(9,1) \text { and at }(4,2)\]

First we check to see that (9,1) satisfies the equation:

\[\sqrt{9}+\sqrt{5-1^{2}}=\sqrt{9}+\sqrt{4}=3+2=5 . \quad \text { It checks. }\]

Then we assume there is a function \(y(x)\) such that

\[\sqrt{x}+\sqrt{5-(y(x))^{2}}=5 \quad \text { and that } \quad y(9)=1\]

We convert the square root symbols to fractional exponents and differentiate using Leibnitz notation.

\[\begin{array}\
x^{\frac{1}{2}}+\left(5-(y(x))^{2}\right)^{\frac{1}{2}} &=5 &\\
\frac{d}{d x}\left(x^{\frac{1}{2}}+\left(5-(y(x))^{2}\right)^{\frac{1}{2}}\right) &=\frac{d}{d x} 5 &\\
\frac{d}{d x} x^{\frac{1}{2}}+\frac{d}{d x}\left(5-(y(x))^{2}\right)^{\frac{1}{2}} &=0 & \text{Sum and Constant Rules}\\
\frac{1}{2} x^{-\frac{1}{2}}+\frac{1}{2}\left(5-y^{2}\right)^{-\frac{1}{2}} \frac{d}{d x}\left(5-(y(x))^{2}\right) &=0 & \text{Power and Power Chain Rules}\\
\frac{1}{2} x^{-\frac{1}{2}}+\frac{1}{2}\left(5-y^{2}\right)^{-\frac{1}{2}}\left(\frac{d}{d x} 5-\frac{d}{d x}(y(x))^{2}\right) &=0 & \text{Sum Rule}\\
\frac{1}{2} x^{-\frac{1}{2}}+\frac{1}{2}\left(5-y^{2}\right)^{-\frac{1}{2}}\left(0-2 y \frac{d}{d x} y(x)\right) &=0 & \text{ Constant and Power Chain Rules}\\
\end{array}\]

Next we solve for \(\frac{d}{dx} y(x)\) and get

\[\frac{d}{d x} y(x)=\frac{\sqrt{5-y^{2}}}{2 y x^{\frac{1}{2}}} \quad \text { and evaluate at }\left.(9,1) \quad \frac{\sqrt{5-y^{2}}}{2 y x^{\frac{1}{2}}}\right|_{(x, y)=(9,1)}=\frac{1}{3}\]

So the slope of the graph at (9,1) is 1/3. You may notice that we have selectively used \(y(x)\) and \(y\); often only \(y\) is used to simplify notation.

An equation of the tangent to the graph of \(\sqrt{x} + \sqrt{5 - y ^{2}} = 5\) at the point (9,1) is \(y - 1 = \frac{1}{3} (x - 9)\).

Now for the point (4,2), the differentiation is exactly the same and we might (alert!) evaluate

\[\frac{d}{d x} y(x)=\frac{\sqrt{5-y^{2}}}{2 y x^{\frac{1}{2}}} \quad \text { at }\left.(4,2) \quad \frac{\sqrt{5-y^{2}}}{2 y x^{\frac{1}{2}}}\right|_{(x, y)=(4,2)}=\frac{1}{8}\]

However, the point (4,2) does not satisfy the original equation and is not a point of its graph. Finding the slope at that point is meaningless, so we punt.

All of this solution is algebraic. The graph of the equation shown in Figure \(\PageIndex{1}\) is of considerable help.

Figure \(\PageIndex{1}\): Solid curve: Graph of \(\sqrt{x}+ \sqrt{5 - y ^{2}} = 5\) and the points (9,1) and (4,2) for Example 4.6.1. Dashed curve: Graph of \(\sqrt{x}+ \sqrt{5 - y ^{2}} = 5\) and the point (36,-2) for Exercise 4.6.2.

 

Exercises for Section 4.6 Implicit Differentiation.

Exercise 4.6.1 For those points that are on the graph, find the slopes of the tangents to the graph of

1. \(\frac{x^{2}}{18} + \frac{y^{2}}{8} = 1\) at the points (3, 2) and (−3, 2)
2. \(\frac{2x ^{2}}{35} + \frac{3y ^{2}}{35} = 1\) at the points (4, 1), (−3, −2), and (4, −1)

Exercise 4.6.2 Find the slope of the graph of \(\sqrt{x}-\sqrt{5-y^{2}}=5\) at the point (36,-2). Is there a slope to the graph at the point (46,1)?

Exercise 4.6.3 A Finnish landscape architect laid out gardens in the shape of the pseudo ellipsoid

\[\frac{|x|^{2.5}}{a^{2.5}}+\frac{|y|^{2.5}}{b^{2.5}}=1\]

a shape that became commonly used in design of Scandinavian furniture and table ware. In Figure Ex. 4.6.3 is the graph of,

\[\frac{|x|^{2.5}}{3^{2.5}}+\frac{|y|^{2.5}}{2^{2.5}}=1\]

and tangents drawn at (2, 1.67) and (-1.5, 1.85). Find the slopes of the tangents.

Figure for Exercise 4.6.3 Graph of the equation \(|x|^{2.5} / 3^{2.5}+|y|^{2.5} / 2^{2.5}=1\) representative of a Scandinavian design.

Exercise 4.6.4 Draw the graph and find the slopes of the tangents to the graph of

1. \(x^{2} - 2y ^{2} = 1\) at the points (3, 2) and (-3, -2)
2. \(2x ^{4} + 3y ^{4} = 35\) at the points (2, 1), (1, -2), and (2, -1)
3. \(\sqrt{|x|} + \sqrt{|y|} = 5\) at the points (9, 4) and (1, -16)
4. \(\sqrt{x} + \sqrt[3] {y} = 9\) at the points (64, 1), (36, 27), and (16, 125)
5. \(x^{2} + y^{2} = (x + y)^{2}\) at the points (1, 0) and (0, 1)
6. \((x ^{2} + 4)y = 24\) at the points (2, 3) and (0, 6)
7. \(x^{3} + y^{3} = (x + y)^{3}\) at the points (1, -1) and (-2, 0)
8. \(x^{4} + x ^{2} y ^{2} = 20y ^{2}\) at the points (2, 1) and (2, -1)

Exercise 4.6.5 This is an interesting and challenging problem. The goal is to explain, among other things, the historical instance of John Adams overhearing the plans of his opponents in Statuary Hall just outside the U.S. congressional chamber.

Ellipses have an interesting reflective property explained by tangents to an ellipse (see Figure \(\PageIndex{2}\)A). Light or sound originating at one focal point of an ellipse is reflected by the ellipse to the other focal point. Statuary Hall is in the shape of an ellipse. John Adams opponents had a desk at one of the focal points and Adams arranged to stand at the other focal point. This property also is a factor in the acoustics of the Mormon Tabernacle in Salt Lake City, Utah and the Smith Civil War Memorial in Philadelphia, Pennsylvania.

Figure \(\PageIndex{2}\): A. An ellipse. Light or sound originating at focal point \(F_1\) and striking the ellipse at (x, y) is reflected to \(F_{2}\). B. The angle of incidence, \(\theta _1\), is equal to the angle of reflection, \(\theta _2\).

In case you have forgotten: For two intersecting lines with inclinations \(\alpha _1\) and \(\alpha _2\) and \(\alpha _{1} > \alpha _{2}\) and slopes \(m_{1} = \tan {\alpha _{1}}\) and \(m_{2} = \tan {\alpha _{2}}\), one of the angles between the two lines is \(\theta = \alpha _{1} - \alpha _{2}\) (Figure \(\PageIndex{3}\)). If neither line is vertical and the lines are not perpendicular,

\[\tan \theta=\tan \left(\alpha_{1}-\alpha_{2}\right)=\frac{\tan \alpha_{1}-\tan \alpha_{2}}{1+\tan \alpha_{1} \tan \alpha_{2}}=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}} \label{4.16}\]

Figure \(\PageIndex{3}\): Two lines with inclinations \(\alpha _{1}\) and \(\alpha _{2} < \alpha _{1}\) and slopes \(m_1\) and \(m_2\). An angle of intersection is \(\theta=\alpha_{1}-\alpha_{2}\).

Refer to Figure \(\PageIndex{2}\)B. Assume an equation of the ellipse is \(b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}, ~ a>b>0\). Then the focal points will be at (-c, 0) and (c, 0) where \(c = \sqrt{a ^{2} - b ^{2}}\). Let \((\bar{x}, \bar{y})\) denote a point of the ellipse. In order to establish the reflective property of ellipses, it is sufficient to show that the angle of incidence, \(\theta _1\) is equal to the angle of reflection, \(\theta _2\).

1. Find the slope, \(m\), of the tangent at \((\bar{x}, \bar{y})\).
2. Show that \[\tan \theta_{1}=\frac{\frac{\bar{y}}{\bar{x}+c}-\left(-\frac{b^{2} \bar{x}}{a^{2} \bar{y}}\right)}{1+\frac{\bar{y}}{\bar{x}+c}\left(-\frac{b^{2} \bar{x}}{a^{2} \bar{y}}\right)}\]
3. Write a similar expression for \(\tan {\theta _2}\).
4. For the algebraically bold. Show that \(\tan {\theta _{1}} = \tan {\theta _{2}}\).
5. Both tangents are positive, both angles are acute, and the angles are equal.