# Implicit differentiation question

1. ### Obsidian

7
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. :)

3. ### Obsidian

7
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?

4. ### awvvu

188
Product rule. You can't just think of y as a constant when you differentiate implicitly.

5. ### Obsidian

7
Yeah, I felt it had something to do with the product rule, but why did they take out the 100 like that?

6. ### awvvu

188
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]

7. ### Obsidian

7
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?

8. ### HallsofIvy

41,265
Staff Emeritus
You really have to know algebra in order to do calculus! 100xy= (100x)y= x(100y)= 100(xy).

9. ### Obsidian

7
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. :/

10. ### rocomath

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

$$\frac{d}{dx}(100x)$$

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