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

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