Calculus/Implicit Differentiation

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Implicit Differentiation

Generally, you will encounter functions expressed in explicit form, that is, in the form . To find the derivative of with respect to , you take the derivative with respect to of both sides of the equation to get

But suppose you have a relation of the form . In this case, it may be inconvenient or even impossible to solve for as a function of . A good example is the relation . In this case you can utilize implicit differentiation to find the derivative. To do so, one takes the derivative of both sides of the equation with respect to and solves for . That is, form

and solve for . You need to employ the chain rule whenever you take the derivative of a variable with respect to a different variable. For example,

Implicit Differentiation and the Chain RuleEdit

To understand how implicit differentiation works and use it effectively it is important to recognize that the key idea is simply the chain rule. First let's recall the chain rule. Suppose we are given two differentiable functions   and that we are interested in computing the derivative of the function   , the chain rule states that:


That is, we take the derivative of   as normal and then plug in   , finally multiply the result by the derivative of   .

Now suppose we want to differentiate a term like   with respect to   where we are thinking of   as a function of   , so for the remainder of this calculation let's write it as   instead of just   . The term   is just the composition of   and   . That is,   . Recalling that   then the chain rule states that:


Of course it is customary to think of   as being a function of   without always writing   , so this calculation usually is just written as


Don't be confused by the fact that we don't yet know what   is, it is some function and often if we are differentiating two quantities that are equal it becomes possible to explicitly solve for   (as we will see in the examples below.) This makes it a very powerful technique for taking derivatives.

Explicit DifferentiationEdit

For example, suppose we are interested in the derivative of   with respect to   , where   are related by the equation


This equation represents a circle of radius 1 centered on the origin. Note that   is not a function of   since it fails the vertical line test (  when   , for example).

To find   , first we can separate variables to get


Taking the square root of both sides we get two separate functions for   :


We can rewrite this as a fractional power:


Using the chain rule we get,


And simplifying by substituting   back into this equation gives


Implicit DifferentiationEdit

Using the same equation


First, differentiate with respect to   on both sides of the equation:


To differentiate the second term on the left hand side of the equation (call it  ), use the chain rule:


So the equation becomes


Separate the variables:


Divide both sides by   , and simplify to get the same result as above:



Implicit differentiation is useful when differentiating an equation that cannot be explicitly differentiated because it is impossible to isolate variables.

For example, consider the equation,


Differentiate both sides of the equation (remember to use the product rule on the term  ):


Isolate terms with  :


Factor out a   and divide both sides by the other term:




can be solved as:


then differentiated:


However, using implicit differentiation it can also be differentiated like this:


use the product rule:


solve for  :


Note that, if we substitute   into   , we end up with   again.

Application: inverse trigonometric functionsEdit

Arcsine, arccosine, arctangent. These are the functions that allow you to determine the angle given the sine, cosine, or tangent of that angle.

First, let us start with the arcsine such that:


To find   we first need to break this down into a form we can work with:


Then we can take the derivative of that:


...and solve for   :

  gives us this unit triangle.

At this point we need to go back to the unit triangle. Since   is the angle and the opposite side is   , the adjacent side is   , and the hypotenuse is 1. Since we have determined the value of   based on the unit triangle, we can substitute it back in to the above equation and get:

Derivative of the Arcsine


We can use an identical procedure for the arccosine and arctangent:

Derivative of the Arccosine


Derivative of the Arctangent


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Implicit Differentiation