# Trouble With Understanding Implicit Differentiation

1. Jul 4, 2011

### nicksbyman

I don't think I fully understand implicit differentiation. I have read my textbook and watched many videos, and I think I will get an A on my test on this solely by memorizing the rules, but I would really like to understand this topic. From what I know, you are supposed to use implicit differentiation when the variable of the thing you are taking the derivative of disagrees with the denominator of d/dx and when they agree use explicit differentiation. I understand that there is a problem with taking the derivative of an expression when the variables disagree, but I don't understand why you can use the chain rule when the variables disagree but you cannot use the simple power rule (i.e. d/dx [yˆ2] = 2y(dy/dx) ≠ 2y). It seems arbitrary; why are you allowed to use the chain rule and not the simple power rule?

Thanks

2. Jul 4, 2011

### lanedance

the difference is you are differentiating y = y(x), a fucntion of x, with respect to x, so you keep the dy/dx term

if you were differentiating with respect to y then the power law applies
$$\frac{d}{dy}y^2 = 2y$$

3. Jul 4, 2011

### lanedance

and in fact you are using both the chain rule and the power rule

4. Jul 4, 2011

### nicksbyman

^I know, but why? Why do you only use the chain rule when the variables disagree?

5. Jul 4, 2011

### QuarkCharmer

I think you actually use the chain rule when the variable agrees. It's just that the derivative of the inside of say X^3 is 1, (that is, the derivative with respect to x of x). So you have something like this:
$$\frac{d}{dx}x^{3}=3x^{2}(\frac{d}{dx}x)$$
The latter part is just one, so we don't have to write it.

Think of it like a composition:
$$f(x) = x^{3}$$
$$g(x) = x$$
$$(f \circ g)(x) = f(g(x)) = (x)^3$$

Last edited: Jul 4, 2011
6. Jul 4, 2011

### lanedance

yeah so just to add to quarkcharmers post

consider the function f(x) = x

then consider g(f) = f^2

taking teh composition of functions we have
$$g(f(x)) = (f(x))^2 = x^2$$

when we differntiate w.r.t. x
$$\frac{d}{dx}g(f(x)) = \frac{dg(f(x))}{dx}g(f(x))\frac{df(x)}{dx} = g'(f(x))f'(x) = 2f(x).1 = 2x$$

so the key part isfor f(x) = x
$$\frac{df(x)}{dx} = f'(x) = \frac{d(x)}{dx} = 1$$

Last edited: Jul 4, 2011
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