# A question about implicit differentiation

1. May 16, 2013

### Dustobusto

So, I understand that implicit differentiation involves derivatives in which x values and y values are mixed up. I've done several implicit differentiation problems a couple sections ago for my math homework, but I pretty just memorized patterns and solved it that way.

Now that I'm trying to make sense of related rates, I think it would help to have a better understanding of some of the reasons why its done the way its done.

So for example, (d/dx) x2 = 2x. that's pretty understandable.

When you take the derivative of y2 its basically done the same way. That's that "pattern memorization" I mentioned. What I don't understand is how they prove this by writing

1. (d/dx) y2

2. (d/dy) (dy/dx) y2

3. (d/dy) y2 (dy/dx)

4. 2y (dy/dx)

Understanding that (dy/dx) in step 4 seems to be an important factor in doing related rates, and I seemed to have missed the significance of that.

If you want a better reference for what I'm talking about, I'm watching this video on youtube and not understanding the "metamorphosis" to prove the derivative of y2

Edit:

So I understand the derivative of any constant times y = ky'

So dy/dx is the same as y' which is what I've been using. Maybe when I do related rates I should look at it that way.

Last edited by a moderator: Sep 25, 2014
2. May 16, 2013

### Dick

It's just the chain rule. Which says (f(y(x)))'=f'(y(x))*y'(x). If f(z)=z^2, then f(y(x))=y(x)^2. The derivative of that is f'(y(x))*y'(x), f'(z)=2z, so the whole derivative is 2*y(x)*y'(x).

Last edited by a moderator: Sep 25, 2014
3. May 16, 2013

### Dustobusto

Yeah, it's starting to dawn on me what this all means. I blame the book though :)

4. May 16, 2013

### Dick

Perhaps, writing (d/dy) (dy/dx) y^2 isn't very clear. It makes it look like you should apply the product rule to the terms after (d/dy). You shouldn't. (dy/dx) (d/dy) y^2 would be much better. It's an ambiguity in writing it that way.