# How to do Implicit Differentiation

Implicit Differentiation

When studying A level mathematics at college you uncover many methods of differentiation, coming to the end of my two year course this is the most difficult I have come across although most of the challenge is getting over the initial appearance of the question.

The problem with implicit differentiation is that you have two unknown terms in your function. In all previous units you would have differentiated products, quotients and functions of functions but there was only one unknown variable in them.

e.g. find dy/dx of x^3 + 2x^2 + 3x – 2 = 3x^2 + 4x + 3
Using usual methods

But a standard function, which you might be asked to implicitly differentiate, will look something like this:

Equation (1)

x^2 + y^2 +3y + 10 = 14

You have to differentiate y and x in terms of x, to do this you effectively tackle the x terms in the regular way but when you come too terms in y you differentiate with respect too y and then multiply by dy/dx, if you do this to the above equation (1) you get:

dy/dx => 2x + 2y(dy/dx) + 3(dy/dx) = 0

This differential is equal to zero since the numbers 10 and 14 both differentiate to zero. The next process is to separate the terms and get all of the dy/dx terms on one side of the equation:

dy/dx => 2y(dy/dx) + 3(dy/dx) = -2x

then you can take out factors of dy/dx:

(dy/dx)(2y + 3) = -2x

Now divide away the unknown terms to get dy/dx on its own, this is your final solution:

Dy/dx = -2x / (2y + 3)

This is a very simple example, many questions will require you to use the product rule or quotient rule, with products of both x and y with terms such as 4xy that you will have to differentiate.

Many questions will use examples of lines and curves to find the equations of the normal or tangents using implicit differentiation an example is below:

A curve with equation (1) x^2 + y^2 +3y + 10 = 14, goes through a point A at (3,4) find the equation of the normal at point A

Well we have already differentiated the equation above and got dy/dx = -2x / (2y + 3), now to find the gradient of the curve at point A we must plug in x = 3 and y = 4:

Gradient = dy/dx = -2(3) / (2(4) + 3) = -6/11
So the gradient of the normal is the reciprocal = 11/6

Now using the standard line equation we can find the answer:

(y-y1) = m (x – x1) x1= 3, y1 = 4, m= 11/6
=> (y – 4) = 11/6 (x – 4)
=> 6y = 11x – 68