Implicit Differentiation
This section covers Implicit Differentiation.
If y^{3} = 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 = 3y^{2}
dy
And now use the fact: | dy | = |
1 |
dx | dx/dy |
So we get:
dy = 1
dx 3y^{2}
Method 3
Differentiate term by term and use the chain rule:
y^{3} = x
d | (y^{3}) | = | 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 (y^{3}) × 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 y^{3}, which is what we had in the previous line).
Therefore, 3y^{2} × dy = 1
dx
So dy = 1
dx 3y^{2}
In this example, method (2) is probably the easiest. However, there are cases when the only possible method is (3).
Example
Differentiate x^{2} + y^{2} = 3x, with respect to x.
d | (x^{2}) | + | d | (y^{2}) | = | d | (3x) |
dx | dx | dx |
2x + d (y^{2})×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