Trevor Pythagoras

Implicit differentiation involves differentiating an equation that hasn’t been arranged such that all of one variable, eg y, is on one side and all of the other variable, eg x, is on the other side. For example differentiating the equation with respect to x (ie find dy/dx):

3x² +2y³ + 6 = 3x²y

TO do this you need to remember that the derivative of y with respect to x is dy/dx, hence when you differentiate a function such as

y=3x

you get

dy/dx = 3

Where the right has gone to the derivative of 3x and the left has gone to the derivative of y, dy/dx.

If you want to differentiate more complex terms involving x you can use the chain rule, since y can be written as a function of x.
so if f(x) = g(y) then
f’(x) = dy/dx g’(y)

(or a simple way of doing it is treat any y’s like x’s and stick a dy/dx on the end).

So for example y² differentiated becomes 2ydy/dx

All the other rules like the product rule and quotient rule still apply.
So to finish lets consider our original equation

3x² +2y³ + 6 = 3x²y

This becomes

6x + 6y²dy/dx = 6xy +3x²dy/dx

Which we can re-arrange to get

dy/dx = (6xy-6x)/(6y²-3x²)=(2xy-2x)/(2y²-1x²)

By David Woodford