Sometimes functions are given not in the form y=f(x) but in a more complicated form in which
it is difficult or even impossible to express y explicitly in terms of x. Such functions are called implicit functions. An example is 4y^5 + 5y = -x.It cannot be rearranged for y the subject, even with a calculator. Here we explain how these can be differentiated using implicit differentiation.

Assumed Knowledge: Before beginning this guide, it is assumed you have knowledge in the following fields:

• General Algebra
  
• Differentiation
  
• Product Rule

Let y be a function in terms of x, and f(x) also be a function in terms of x.Then it logically follows that:

(d/dx((f(y)))=(d/dy(f(y)))(dy/dx)

While this result is elementary, it is important in differentiating in terms of some equation y.

The basic idea of implicit differentiation is to take d/dx of each term, and then rearranged for dy/dx.

Question: Find dy/dx of 3xy^2 + 2y = x.

Solution: d/dx(3xy^2) + d/dx(2y) = d/dx(x). Let’s tackle this term by term.

For d/dx(3xy^2), we will need to use the product rule, that (f(x)g(x))'=f(x)g'(x)+f'(x)g(x).Of course, we are dealing with y^2,so we will need to use implicit differentiation:

d/dx(3xy^2) = (3x)(d/dx(y2^))+3y^2

Recall the rule that d/dx(f(y))=d/dy(f(y))(dy/dx):

d/dx(3xy^2) = (3x)(d/dy((y^2)))(dy/dx)
d/dx(3xy^2) = (6xy)(dy/dx) + (3y^2)

Now the second term, (d/dx((2y))).This is simply 2(dy/dx).

d/dx(x) = 1, obviously.

So we have (6xy)(dy/dx) + 2(dy/dx) + 3y^2 = 1.We now rearrange for dy/dx:

dy/dx(6xy + 2) = 1 - 3y^2
dy/dx=(1 - 3y^2)/(6xy + 2)

Remarks:
In general testing of this example with 30 people, this question was not well answered. Many people tried to rearrange for y when this is not possible. Others made calculation errors. Some did not correctly apply the product rule when this is required in order to answer the question.

In this guide, we discussed how to do basic implicit differentiation. Implicit differentiation is useful when the equation cannot be rearranged for y the subject. By using implicit differentiation, we can extend applications of calculus to find the equation of the tangent to the curve beyond simple polynomial functions.