# Implicit differentiation

Source: https://projectalevel.co.uk/maths/calculus/implicit_differentiation

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

1) rewrite it as y = x to the power of 1/3 and differentiate as normal.

2) dx = 3y²
 dy

Since dy = 1
 dx dx/dy

dy = 1
dx 3y²

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 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 clearly 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 (y²) × dy = 3
 dy dx

2x + 2y dy = 3
 dx

dy = 3 - 2x
dx 2y