Calculus/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 Rule
[edit  edit source]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 Differentiation
[edit  edit source]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 Differentiation
[edit  edit source]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:
Uses
[edit  edit source]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:
Example
[edit  edit source]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 functions
[edit  edit source]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 :
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:

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

