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
