# Implicit differentiation

1. Jul 9, 2008

### jgens

Could someone please explain to me how implicit differentiation is an application of the chain rule? It would be much appreciated. By the way, if it helps, I'm a junior in high school. Thanks.

2. Jul 9, 2008

### epkid08

Let's say y=(x+2)^2. We would most likely substitute u in for (x+2), and then say the derivative is the derivative of u^2; you would simplify that using the chain rule. Well, in that case, u, is a function of x, u(x)=(x+2). Implicit differentiation is just when y is not by itself, and possibly not solvable to be by itself. Example: y^2=x. Y is a function of x, so you have to treat y as you treated U in the last equation; Simplify using the chain rule.

3. Jul 9, 2008

### cristo

Staff Emeritus
Check out the library article on the subject (click on the underlined phrase 'implicit differentiation' in the previous post)

4. Jul 10, 2008

### arildno

A simple example is x-y=0
We see that y=Y(x)=x, so that dy/dx=1

Now, set g(x,y)=x-y. Clearly h(x)=g(x,Y(x))=x-x=0 IDENTICALLY, for all values of x!

Thus, we may differentiate h(x)=0, since this "equation" holds for all x's.
We get:
$$\frac{dh}{dx}=\frac{\partial{g}}{\partial{x}}+\frac{\partial{g}}{\partial{y}}\frac{dY}{dx}=0.$$
We have:
$$\frac{\partial{g}}{\partial{x}}=1,\frac{\partial{g}}{\partial{y}}=-1$$, and therefore:
$$1-1\frac{dY}{dx}=0\to\frac{dY}{dx}=1$$, as we should have.

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