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,
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.
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: