Premed · Premed · Calculus 1
Lecture 9: Implicit Differentiation
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Distinguish between explicit and implicit definitions of functions
- Differentiate equations implicitly to find dy/dx
- Find tangent lines to implicitly defined curves
- Apply implicit differentiation to find higher-order derivatives
- Differentiate inverse trigonometric functions using implicit differentiation
Lecture Content
I. Explicit vs. Implicit Functions
In explicit form, y is isolated on one side of the equation, as in y = x^2 + 3x. In implicit form, x and y are intertwined in an equation without y being solved for explicitly. For example, x^2 + y^2 = 25 defines a circle, and x^3 + y^3 = 6xy defines the folium of Descartes. Neither of these is a function in the strict sense, but each defines y implicitly as a function of x on portions of the curve.
Sometimes it is difficult or outright impossible to solve explicitly for y. Implicit differentiation provides a way to find dy/dx without ever isolating y.
II. The Method of Implicit Differentiation
The procedure is straightforward. First, differentiate both sides of the equation with respect to x. Second, treat y as a function of x, so every time you differentiate a term involving y, apply the chain rule and multiply by dy/dx. Third, collect all terms containing dy/dx on one side. Fourth, solve for dy/dx.
For the circle x^2 + y^2 = 25, differentiating gives 2x + 2y(dy/dx) = 0, which yields dy/dx = -x/y. At the point (3, 4), the slope is dy/dx = -3/4.
For the folium x^3 + y^3 = 6xy, differentiating gives 3x^2 + 3y^2(dy/dx) = 6y + 6x(dy/dx). Rearranging, 3y^2(dy/dx) - 6x(dy/dx) = 6y - 3x^2, and solving for the derivative yields dy/dx = (2y - x^2)/(y^2 - 2x).
<image>The folium of Descartes, defined by x^3 + y^3 = 6xy, plotted as a curve with a loop in the first quadrant. A point on the curve is marked, with the tangent line drawn through it. The implicit derivative formula dy/dx = (2y - x^2)/(y^2 - 2x) is displayed next to the curve. The axes and asymptote line x + y = -2 are also shown. Title: "Implicit differentiation applied to the folium of Descartes."</image>
III. Tangent and Normal Lines to Implicit Curves
Once dy/dx is found, the tangent line at a point (a, b) on the curve is y - b = (dy/dx)|_{(a,b)} (x - a), and the normal line (perpendicular to the tangent) is y - b = -1/(dy/dx)|_{(a,b)} (x - a). It is important to verify that the point actually lies on the curve by substituting into the original equation.
For x^2 + y^2 = 25 at the point (3, 4), the tangent line is y - 4 = (-3/4)(x - 3), or equivalently 3x + 4y = 25. The normal line is y - 4 = (4/3)(x - 3).
IV. Higher-Order Implicit Derivatives
To find d^2y/dx^2, differentiate dy/dx implicitly once more. Starting from x^2 + y^2 = 25, we found dy/dx = -x/y. Applying the quotient rule and implicit differentiation gives d^2y/dx^2 = d/dx[-x/y] = [-y - (-x)(dy/dx)] / y^2 = [-y + x(-x/y)] / y^2 = [-y^2 - x^2] / y^3 = -25/y^3, where the last step uses the original equation x^2 + y^2 = 25 to simplify.
V. Derivatives of Inverse Trigonometric Functions
Implicit differentiation provides an elegant way to derive the derivatives of inverse trigonometric functions. For arcsin x, let y = arcsin x, so sin y = x. Differentiating implicitly gives cos y * (dy/dx) = 1, hence dy/dx = 1/cos y = 1/sqrt(1 - sin^2 y) = 1/sqrt(1 - x^2). Thus d/dx [arcsin x] = 1/sqrt(1 - x^2) for -1 < x < 1. By a similar argument, d/dx [arccos x] = -1/sqrt(1 - x^2).
For arctan x, let y = arctan x, so tan y = x. Differentiating gives sec^2 y * (dy/dx) = 1, so dy/dx = 1/sec^2 y = 1/(1 + tan^2 y) = 1/(1 + x^2). Thus d/dx [arctan x] = 1/(1 + x^2).
The remaining inverse trig derivatives are d/dx [arccot x] = -1/(1 + x^2), d/dx [arcsec x] = 1/(|x| sqrt(x^2 - 1)) for |x| > 1, and d/dx [arccsc x] = -1/(|x| sqrt(x^2 - 1)) for |x| > 1.
<image>A reference table of inverse trigonometric derivatives in two columns. Left column: function (arcsin x, arccos x, arctan x, arccot x, arcsec x, arccsc x). Right column: derivative with domain restrictions noted. Below the table, a right triangle diagram shows how sin y = x leads to cos y = sqrt(1 - x^2), which is the key step in deriving d/dx[arcsin x]. The triangle has hypotenuse 1, opposite side x, and adjacent side sqrt(1 - x^2). Title: "Inverse Trig Derivatives via the Reference Triangle."</image>
VI. Chain Rule with Inverse Trig Functions
When inverse trig functions have composite arguments, the chain rule applies in the usual way. The formulas become d/dx [arctan(g(x))] = g'(x) / (1 + [g(x)]^2) and d/dx [arcsin(g(x))] = g'(x) / sqrt(1 - [g(x)]^2).
For example, d/dx [arctan(3x)] = 3/(1 + 9x^2), and d/dx [arcsin(x^2)] = 2x / sqrt(1 - x^4). For d/dx [x * arctan x], the product rule combined with the inverse trig derivative gives arctan x + x/(1 + x^2).
VII. Implicit Differentiation in Related Contexts
Implicit differentiation is not limited to algebraic curves. It can be applied whenever y is defined implicitly through any equation involving x and y, including equations with exponentials, logarithms, and trigonometric functions.
For instance, given e^y + xy = 1, differentiating both sides yields e^y(dy/dx) + y + x(dy/dx) = 0. Factoring gives dy/dx(e^y + x) = -y, so dy/dx = -y/(e^y + x). Similar techniques apply to equations like sin(xy) = x or ln y + y = x^2.
