This section covers Implicit Differentiation.
If y3 = x, how would you differentiate this with respect to x? There are three ways:
Rewrite it as y = x(1/3) and differentiate as normal (in harder cases, this is not possible!)
dx = 3y2
|And now use the fact:||dy||=||
So we get:
dy = 1
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
(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
So dy = 1
In this example, method (2) is probably the easiest. However, there are cases when the only possible method is (3).
Differentiate x2 + y2 = 3x, with respect to x.
2x + d (y2)×dy = 3
2x + 2y dy = 3
dy = 3 - 2x
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