Implicit differentiation question

Obsidian

Find y' = dy/dx for x3 + y3 = 4


Okay, now what's really confusing me is that for the y3 is that you need to use the chain rule for it. When you do, the answer is 3y2(dy/dx). How does that actually work?

And if anyone can give me any good advice on any good guidelines on how to properly implicitly differentiate, it would be most helpful. Thanks. :)

rocomath

http://en.wikipedia.org/wiki/Implici...ifferentiation

[tex]3x^2+3y^2\frac{dy}{dx}=0[/tex]

Obsidian

My next question is, in my book, they differentiate d/dx(100xy) and turn it into:

100[x{dy/dx) + y]

How did they get to that?

awvvu

Product rule. You can't just think of y as a constant when you differentiate implicitly.

Obsidian

Yeah, I felt it had something to do with the product rule, but why did they take out the 100 like that?

awvvu

It's just a constant, so that can be moved out. d/dx[100xy] = 100 d/dx[xy] = 100[x*dy/dx + dx/dx*y]

Obsidian

But can't you do that only if the 100 is being multiplied in both x and y? I mean, the way it is, wouldn't it only be multiplying to either the X or the y, not twice to both of them?

HallsofIvy

You really have to know algebra in order to do calculus! 100xy= (100x)y= x(100y)= 100(xy).

Obsidian

No, I know that's how it works, but I guess my question is, why isn't the derivative of 100 being taken as well? Why is it being left out?

Sorry for the dumb questions. :/

rocomath

Ok, take the derivative of this problem. Treat it as if it's a product.

[tex]\frac{d}{dx}(100x)[/tex]

What is your answer?

Now go back to your derivative properties in which, [tex]\frac{d}{dx}cf(x)=cf'(x)[/tex]

Now do you see why it can be left out?

Obsidian

Ah, of course! Me so dumb. Thanks a lot. :)