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  • Find the derivative of a complicated function by using implicit differentiation.
  • Use implicit differentiation to determine the equation of a tangent line.

We have already studied how to find equations of tangent lines to functions and the rate of change of a function at a specific point. In all these cases we had the explicit equation for the function and differentiated these functions explicitly. Suppose instead that we want to determine the equation of a tangent line to an arbitrary curve or the rate of change of an arbitrary curve at a point. In this section, we solve these problems by finding the derivatives of functions that define y implicitly in terms of x .

Implicit differentiation

In most discussions of math, if the dependent variable y is a function of the independent variable x , we express y in terms of x . If this is the case, we say that y is an explicit function of x . For example, when we write the equation y = x 2 + 1 , we are defining y explicitly in terms of x . On the other hand, if the relationship between the function y and the variable x is expressed by an equation where y is not expressed entirely in terms of x , we say that the equation defines y implicitly in terms of x . For example, the equation y x 2 = 1 defines the function y = x 2 + 1 implicitly.

Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test). We are using the idea that portions of y are functions that satisfy the given equation, but that y is not actually a function of x .

In general, an equation defines a function implicitly if the function satisfies that equation. An equation may define many different functions implicitly. For example, the functions

y = 25 x 2 and y = { 25 x 2 if 25 x < 0 25 x 2 if 0 x 25 , which are illustrated in [link] , are just three of the many functions defined implicitly by the equation x 2 + y 2 = 25 .

The circle with radius 5 and center at the origin is graphed fully in one picture. Then, only its segments in quadrants I and II are graphed. Then, only its segments in quadrants III and IV are graphed. Lastly, only its segments in quadrants II and IV are graphed.
The equation x 2 + y 2 = 25 defines many functions implicitly.

If we want to find the slope of the line tangent to the graph of x 2 + y 2 = 25 at the point ( 3 , 4 ) , we could evaluate the derivative of the function y = 25 x 2 at x = 3 . On the other hand, if we want the slope of the tangent line at the point ( 3 , −4 ) , we could use the derivative of y = 25 x 2 . However, it is not always easy to solve for a function defined implicitly by an equation. Fortunately, the technique of implicit differentiation    allows us to find the derivative of an implicitly defined function without ever solving for the function explicitly. The process of finding d y d x using implicit differentiation is described in the following problem-solving strategy.

Problem-solving strategy: implicit differentiation

To perform implicit differentiation on an equation that defines a function y implicitly in terms of a variable x , use the following steps:

  1. Take the derivative of both sides of the equation. Keep in mind that y is a function of x . Consequently, whereas d d x ( sin x ) = cos x , d d x ( sin y ) = cos y d y d x because we must use the chain rule to differentiate sin y with respect to x .
  2. Rewrite the equation so that all terms containing d y d x are on the left and all terms that do not contain d y d x are on the right.
  3. Factor out d y d x on the left.
  4. Solve for d y d x by dividing both sides of the equation by an appropriate algebraic expression.

Questions & Answers

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Practice Key Terms 1

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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