# Implicit Differentiation

This section covers Implicit Differentiation.

If y3 = x, how would you differentiate this with respect to x? There are three ways:

Method 1

Rewrite it as y = x(1/3) and differentiate as normal (in harder cases, this is not possible!)

Method 2

Find dx/dy:

dx  =  3y2

dy

 And now use the fact: dy = 1 dx dx/dy

So we get:

dy  =  1

dx     3y2

Method 3

Differentiate term by term and use the chain rule:

y3   =   x

 d (y3) = d (x) dx dx

The right hand side of this equation is 1, since the derivative of x is 1. However, to work out the left hand side we must use the chain rule.

The left hand side becomes:

d (y3) ×  dy

dy         dx

(although it is not strictly correct to do so, at this level you can think of dy/dx as a fraction in the chain rule. In the line above, imagine that you can cancel the "dy" s, leaving d/dx and y3, which is what we had in the previous line).

Therefore, 3y2 × dy  = 1

dx

So   dy  =  1

dx    3y2

In this example, method (2) is probably the easiest. However, there are cases when the only possible method is (3).

Example

Differentiate x2 + y2 = 3x, with respect to x.

 d (x2) + d (y2) = d (3x) dx dx dx

2x + d (y2dy  =  3

dy        dx

2x + 2y dy  = 3

dx

dy  =  3 - 2x

dx        2y

Example

Differentiate ax with respect to x.

You might be tempted to write xax-1  as the answer. This is wrong. That would be the answer if we were differentiating with respect to a not x.

Put y = ax .

Then, taking logarithms of both sides, we get:

ln y = ln (ax)

so ln y = x lna

So, differentiating implicitly, we get: (1/y) (dy/dx) = lna

and so dy/dx = y lna = ax lna