# Implicit differentiation

Up a level : Differentiation, derivatives
Next page : dx/dy in three different ways Say you have a mathematical relation of two variables, the independent x and the dependent y where you don’t know how to make y the subject, but you still need the derivative dy/dx. What to do? Happily enough there is a method, and that is to realize that, y can be seen as a function of x, and then we use the chain rule. So we get $(f(y))' = f'(y)y'$

Let us put this to a very simple practice where we can check the result.  Let us say that ${y^2} = x,\quad x \geqslant 0$

so that $y = \sqrt x$

and thus $\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}$

If we instead take the derivative of the original relation directly we get $2yy' = 1$

or $y' = \frac{1}{{2y}}$

But $y = \sqrt x$

so $y' = \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}$

We will look at a few examples where it either gives insights to relations we technically could make the dependent variable the subject the derivative of, or it enable us to find properties of relations where we cannot make the dependent variable the subject. Up a level : Differentiation, derivatives
Next page : dx/dy in three different ways Last modified: Feb 16, 2018 @ 10:12