D - Implicit Differentiation Lesson

Implicit Differentiation

Explicitly Defined and Implicitly Defined Functions

An explicit function defines a relationship that equates the dependent variable directly with a function of the independent variable, as in y = f(x), so that its values may be directly calculated from those of the independent variable. In contrast an implicit function is a relationship that does not express the value of the dependent variable directly as a function of the independent variable (y = f(x)), but rather states a relationship which the variables must jointly satisfy (F(x, y) = 0).

Many equations of the form F(x, y) = 0 that define relationships between x and y are not easily solved for y explicitly in terms of x. In most cases the graph of F(x, y) = 0 over its domain fails the vertical line test for functions. However, various sections of the graph may well be functions of x and these sections are implicitly defined by F(x, y) = 0. Under these circumstances, we can then describe y as being a differentiable function of x, and we can find dy/dx using the technique of implicit differentiation.

graph of semi circle 1 graph of semi circle 2

graph of semi circle 3 graph of semi circle 4

Steps for Implicit Differentiation

It is not necessary to solve an equation for y in terms of x in order to find the derivative of y. The method of implicit differentiation can be used instead.

Differentiate both sides of the equation with respect to x.
Collect the terms with dy/dx on one side of the equation.
Factor out dy/dx.
Solve for dy/dx by dividing.

View the presentation below on implicit differentiation from the beginning to 11:30.

Implicit differentiation involving trigonometric functions and the Chain Rule are explored in the presentations below.

Equations of Tangent and Normal Lines for Implicitly Defined Functions

One of the most famous problems involving a tangent line drawn to an implicitly defined function is the folium of Descartes. In general, this curve is defined by x³ + y³ - 3axy = 0 and has an asymptote of x + y + a = 0.

Click HERE to view a presentation on finding the equation of the tangent line to the folium x³+y³=6xy.

The video below is a presentation on finding a tangent line via implicit differentiation.

Did You Know? The word normal comes from the Latin word normalis, which scholars used for perpendicular. A line perpendicular to a tangent line to a curve at the point where the tangent line intersects the curve is a normal line. If the slope of a tangent line is m, then the slope of the normal line is -1/m. The normal line to the folium of Descartes has a slope of 1 and the normal line equation is y = x.

Implicit Differentiation Practice

1. Find the slope of the tangent line to x²y - y²x = -1 at (1, 1).

Solution: 1 text annotation indicator

2. Find y" by implicit differentiation for x³y³ = 5.

Solution: $LaTeX: y^n=\frac{2y}{x^2}$ $y^n=\frac{2y}{x^2}$ text annotation indicator

3. Find equation image indicator $LaTeX: \frac{dy}{dx}$ $\frac{dy}{dx}$ if $LaTeX: \tan^3\left(x^2-y^2\right)=\tan\left(\frac{\pi}{4}\right)$ $\tan^3\left(x^2-y^2\right)=\tan\left(\frac{\pi}{4}\right)$ .

Solution: $LaTeX: \frac{dy}{dx}=\frac{x}{y\:}\:or\:\frac{x}{y}$ $\frac{dy}{dx}=\frac{x}{y\:}\:or\:\frac{x}{y}$ text annotation indicator

4. Find the slope of the normal line to x² - 4y² = 5 at (3, 1).

Solution: text annotation indicator -4/3

5. Find the equations of the vertical tangent lines to the ellipse 5x ² - 6xy + 5y ² = 16.

Solution text annotation indicator : $LaTeX: x=\sqrt[]{5}\:\&\:x=-\sqrt[]{5}$ $x=\sqrt[]{5}\:\&\:x=-\sqrt[]{5}$

Implicit Differentiation: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of implicit differentiation.

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