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Lecture 13: Implicit Differentiation

GOAL: Differentiate functions of the form \( F(x,y) = 0 \). Use this to find the derivatives of inverse trigonometric functions.

Recall the following:

  1. If \( y \) is a function of \( x \), then by the Chain Rule:

    \[ \begin{gathered} \frac{d}{dx}\left(y^2\right) = \frac{d}{dx}\left(\left[y(x)\right]^2\right) = 2y \cdot \frac{dy}{dx} \\[0.5em] \frac{d}{dx}\left(\sin(y)\right) = \frac{d}{dx}\left(\sin(y(x))\right) = \cos y \cdot \frac{dy}{dx} \\[0.5em] \frac{d}{dx}\left(\ln(y)\right) = \frac{d}{dx}\left(\ln(y(x))\right) = \frac{1}{y} \cdot \frac{dy}{dx} \end{gathered} \]

  2. Taking the derivative does not change equality.

    Take the function \( y = 2x+1 \) as an example. We know \( y' = 2 \), but we can also do the following:

    \[ \begin{array}{rcl} y - 2x - 1 &=& 0 \\ \frac{d}{dx}\left(y - 2x - 1\right) &=& \frac{d}{dx}(0) \\ y' - 2 &=& 0 \\ y' &=& 2 \end{array} \]

Motivating Question How can we find the equation of a line tangent to the unit circle \( x^2 + y^2 = 1 \)?

Answer: Recognize that \( y \) is a function of \( x \).

Unit circle on x-y axes with a red tangent line touching a point in the second quadrant, labeled x squared plus y squared equals 1.
Figure 1
A hand-drawn sketch inside a rectangular box. A pair of perpendicular axes with arrowheads at all four ends is centered in the box, and a circle of radius 1 is drawn centered at the origin, crossing each axis. A small red dot marks a point on the circle in the upper-left (second quadrant) region, and a short red straight line passes through that dot, slanting downward from upper-left to lower-right, representing the tangent line to the circle at that point. Below the axes, inside the box, the equation \( x^2 + y^2 = 1 \) is written as the label for the circle.
\[ \begin{array}{rcl} x^2 + y^2 &=& 1 \\ \frac{d}{dx}\left[x^2 + y^2\right] &=& \frac{d}{dx}(1) \\ 2x + 2y\frac{dy}{dx} &=& 0 \\ 2y\frac{dy}{dx} &=& -2x \\ \frac{dy}{dx} &=& \frac{-2x}{2y} \\ \frac{dy}{dx} &=& -\frac{x}{y} \end{array} \]

So, for any point \( (x,y) \) on the unit circle, we can find the slope of the tangent to the unit circle at the point \( (x,y) \).

Notice that \( \frac{dy}{dx} \) is a function of both \( x \) and \( y \); usually, this is the best we can do. However, we can make \( \frac{dy}{dx} \) solely a function of \( x \) in this particular example.

\[ \begin{array}{rcl} x^2 + y^2 &=& 1 \\ y^2 &=& 1 - x^2 \\ y &=& \pm\sqrt{1-x^2} \end{array} \]

Hence,

\[ \frac{dy}{dx} = -\frac{x}{y} = -\frac{x}{\pm\sqrt{1-x^2}} \]

Breaking into cases we get

\[ \frac{dy}{dx} = \begin{cases} -\dfrac{x}{\sqrt{1-x^2}} & \text{if } y > 0 \\[2ex] \dfrac{x}{\sqrt{1-x^2}} & \text{if } y < 0 \end{cases} \]

\( x^2 + y^2 = 1 \) is an example of an implicit function. Finding \( \frac{dy}{dx} \) is called implicit differentiation.

Examples

Explicit Functions and Implicit Functions
Explicit Functions Implicit Functions
\( y = f(x) \) \( F(x,y) = 0 \)
\( y = 2x - 1 \) \( x^2 + y^2 - 1 = 0 \)
\( y = \frac{1}{x} \) \( xy - 1 = 0 \)
\( y = e^{x^2} \) \( e^{xy} + x^2 = 0 \)
\( y = \sin(x) \) \( y\sin(xy) - \pi = 0 \)

Example 1

Find the slope of the line tangent to the elliptic curve

\[ y^2 = x^3 - x + 1 \]

at \( (x,y) = (0,1) \).

NOTE:

If we want to be proper, the first step is the verify \( (0,1) \) is actually on the curve. Replacing \( x \) with 0 and \( y \) with 1 we get

\[ \begin{array}{rcl} 1^2 &=& 0^3 - 0 + 1 \\ 1 &=& 1 \ \checkmark \end{array} \]

For our purposes, you may assume the point given to you lies on the curve.

Hand-drawn elliptic curve with an oval loop and a branch, with a blue tangent line at the point (0,1).
Figure 2
A freehand sketch in red of the elliptic curve \( y^2 = x^3 - x + 1 \) on x and y axes. The curve consists of a closed oval loop on the left, crossing the x-axis at approximately \( x = -1.3 \) (labeled \( \approx -1.3 \)), and an open branch on the right that passes through the points \( (0,1) \) and \( (0,-1) \) on the y-axis, marked with small dots and labeled 1 and -1. A blue straight line is drawn through the marked point \( (0,1) \), sloping gently downward from left to right, representing the tangent line with slope \( -\frac{1}{2} \).
\[ \begin{array}{rcl} y^2 &=& x^3 - x + 1 \\ \frac{d}{dx}\left(y^2\right) &=& \frac{d}{dx}\left(x^3 - x + 1\right) \\ 2y\frac{dy}{dx} &=& 3x^2 - 1 \\ \frac{dy}{dx} &=& \frac{3x^2 - 1}{2y} \\ \left.\frac{dy}{dx}\right]_{(x,y)=(0,1)} &=& \frac{3(0)^2 - 1}{2(1)} \\ \left.\frac{dy}{dx}\right]_{(x,y)=(0,1)} &=& -\frac{1}{2} \end{array} \]

Example 2

Find \( \frac{dy}{dx} \) for the algebraic curve

\[ y^3 + yx^2 = x^3 + xy^2 \]

We start by taking the derivative of both sides.

\[ \begin{array}{rcl} y^3 + yx^2 &=& x^3 + xy^2 \\ \frac{d}{dx}\left(y^3 + yx^2\right) &=& \frac{d}{dx}\left(x^3 + xy^2\right) \\ 3y^2\frac{dy}{dx} + \left[\frac{dy}{dx}x^2 + y(2x)\right] &=& 3x^2 + \left[(1)y^2 + x\left(2y\frac{dy}{dx}\right)\right] \\ 3y^2\frac{dy}{dx} + x^2\frac{dy}{dx} + 2xy &=& 3x^2 + y^2 + 2xy\frac{dy}{dx} \end{array} \]

The next step is to bring every term with a \( \frac{dy}{dx} \) in it to the left-hand side, and the remaining terms to the right-hand side.

\[ \begin{array}{rcl} 3y^2\frac{dy}{dx} + x^2\frac{dy}{dx} + 2xy &=& 3x^2 + y^2 + 2xy\frac{dy}{dx} \\ 3y^2\frac{dy}{dx} + x^2\frac{dy}{dx} - 2xy\frac{dy}{dx} &=& 3x^2 + y^2 - 2xy \end{array} \]

Factor out \( \frac{dy}{dx} \) and solve for \( \frac{dy}{dx} \).

\[ \begin{array}{rcl} 3y^2\frac{dy}{dx} + x^2\frac{dy}{dx} - 2xy\frac{dy}{dx} &=& 3x^2 + y^2 - 2xy \\ \left(3y^2 + x^2 - 2xy\right)\frac{dy}{dx} &=& 3x^2 + y^2 - 2xy \\ \frac{dy}{dx} &=& \frac{3x^2 + y^2 - 2xy}{3y^2 + x^2 - 2xy} \end{array} \]

NOTE:

We can use this strategy to deal with basically any Implicit Differentiation problem.

Example 3

Find \( y' \) given \( xe^y = 9y \)

\[ \begin{array}{rcl} xe^y &=& 9y \\ \left[xe^y\right]' &=& \left[9y\right]' \\ (1)e^y + xe^y y' &=& 9y' \\ xe^y y' - 9y' &=& -e^y \\ \left[xe^y - 9\right]y' &=& -e^y \\ y' &=& \frac{-e^y}{xe^y - 9} \\ y' &=& \frac{e^y}{9 - xe^y} \end{array} \]

Example 4

Find \( \frac{dy}{dx} \) given \( 10\sin(2x + 3y) = 11xy \)

\[ \begin{array}{rcl} 10\sin(2x + 3y) &=& 11xy \\ \frac{d}{dx}\left[10\sin(2x + 3y)\right] &=& \frac{d}{dx}\left[11xy\right] \\ 10\cos(2x + 3y)\left[2 + 3\frac{dy}{dx}\right] &=& 11y + 11x\frac{dy}{dx} \\ 20\cos(2x + 3y) + 30\cos(2x + 3y)\frac{dy}{dx} &=& 11y + 11x\frac{dy}{dx} \\ 30\cos(2x + 3y)\frac{dy}{dx} - 11x\frac{dy}{dx} &=& 11y - 20\cos(2x + 3y) \\ \left[30\cos(2x + 3y) - 11x\right]\frac{dy}{dx} &=& 11y - 20\cos(2x + 3y) \end{array} \]\[ \frac{dy}{dx} = \frac{11y - 20\cos(2x + 3y)}{30\cos(2x + 3y) - 11x} \]

Example 5

Find \( \frac{dy}{dx} \) given \( \frac{1}{x} + \frac{1}{y} = 1 \)

\[ \begin{array}{rcl} \frac{1}{x} + \frac{1}{y} &=& 1 \\ \frac{d}{dx}\left[\frac{1}{x} + \frac{1}{y}\right] &=& \frac{d}{dx}(1) \\ -\frac{1}{x^2} - \frac{1}{y^2}\frac{dy}{dx} &=& 0 \\ \frac{1}{y^2}\frac{dy}{dx} &=& -\frac{1}{x^2} \\ \frac{dy}{dx} &=& -\frac{y^2}{x^2} \end{array} \]

Example 6

Find \( \frac{dy}{dx} \) if \( \sqrt{x} + \sqrt{y} = 4 \)

\[ \begin{array}{rcl} \sqrt{x} + \sqrt{y} &=& 4 \\ \frac{d}{dx}\left[\sqrt{x} + \sqrt{y}\right] &=& \frac{d}{dx}\left[4\right] \\ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} &=& 0 \\ \frac{2\sqrt{y}}{2\sqrt{x}} + \frac{dy}{dx} &=& 0 \\ \frac{dy}{dx} &=& -\frac{\sqrt{y}}{\sqrt{x}} \\ \frac{dy}{dx} &=& -\sqrt{\frac{y}{x}} \end{array} \]

Derivatives of Inverse Sine and Tangent

Implicit Differentiation is particularly helpful if we need to find the derivative of a differentiable function's inverse.

Example 7

Find \( \frac{dy}{dx} \) given \( y = \sin^{-1}(x) \)

\[ \begin{array}{rcl} y &=& \sin^{-1}(x) \\ \sin(y) &=& x \\ \frac{d}{dx}\left[\sin(y)\right] &=& \frac{d}{dx}\left[x\right] \\ \cos(y)\frac{dy}{dx} &=& 1 \\ \frac{dy}{dx} &=& \frac{1}{\cos(y)} \\ \frac{dy}{dx} &=& \frac{1}{\cos\left(\sin^{-1}(x)\right)} \end{array} \]

NOTE:

In general, an inverse trig function inside of a trig function is not considered simplified. To simplify, let \( \theta = \sin^{-1} x \). Then, \( \frac{x}{1} = \sin\theta \), which we can use to make a right triangle.

Right triangle with angle theta, opposite side x, hypotenuse 1, adjacent side square root of 1 minus x squared
Figure 3
A hand-drawn right triangle in blue ink. The hypotenuse slopes upward from the lower-left vertex to the upper-right and is labeled 1. The vertical right side on the right is labeled x. The horizontal bottom side is labeled \( \sqrt{1-x^2} \). The angle at the lower-left vertex, between the hypotenuse and the horizontal side, is marked \( \theta \). The right angle is at the lower-right vertex where the horizontal and vertical sides meet.

Therefore,

\[ \begin{array}{rcl} \frac{dy}{dx} &=& \frac{1}{\cos\left(\sin^{-1}(x)\right)} \\ \frac{dy}{dx} &=& \frac{1}{\cos(\theta)} \\ \frac{dy}{dx} &=& \frac{1}{\sqrt{1-x^2}} \end{array} \]

Example 8

Find \( \frac{dy}{dx} \) given \( y = \tan^{-1}(x) \)

\[ \begin{array}{rcl} y &=& \tan^{-1}(x) \\ \tan(y) &=& x \\ \frac{d}{dx}\left[\tan(y)\right] &=& \frac{d}{dx}\left[x\right] \\ \sec^2(y)\frac{dy}{dx} &=& 1 \\ \frac{dy}{dx} &=& \frac{1}{\sec^2(y)} \\ \frac{dy}{dx} &=& \frac{1}{\left[\sec\left(\tan^{-1}(x)\right)\right]^2} \end{array} \]

Again, if \( \theta = \tan^{-1}x \). Then we can make a right triangle

Right triangle with angle theta, opposite side x, adjacent side 1, hypotenuse square root of x squared plus 1
Figure 4
A hand-drawn right triangle. The angle \( \theta \) is at the lower-left vertex. The horizontal bottom side, adjacent to \( \theta \), is labeled 1. The vertical right side, opposite \( \theta \), is labeled \( x \), and a small square marks the right angle between the bottom and right sides. The slanted hypotenuse running from the lower-left vertex up to the top-right vertex is labeled \( \sqrt{x^2+1} \).

So,

\[ \begin{array}{rcl} \frac{dy}{dx} &=& \frac{1}{\left[\sec(\theta)\right]^2} \\ \frac{dy}{dx} &=& \frac{1}{1+x^2} \end{array} \]