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

So we get:

dy  =  1

dx     3y²

Method 3

Differentiate term by term and use the chain rule:

y³   =   x

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.

2x + d (y²)×dy  =  3

       dy        dx

2x + 2y dy  = 3

           dx

dy  =  3 - 2x

dx        2y

Example

Differentiate a^x with respect to x.

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

Put y = a^x .

Then, taking logarithms of both sides, we get:

ln y = ln (a^x)

so ln y = x lna

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

and so dy/dx = y lna = a^x lna