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Implicit differentiation question

  1. Feb 18, 2008 #1
    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. :)
  2. jcsd
  3. Feb 18, 2008 #2
  4. Feb 18, 2008 #3
    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?
  5. Feb 18, 2008 #4
    Product rule. You can't just think of y as a constant when you differentiate implicitly.
  6. Feb 18, 2008 #5
    Yeah, I felt it had something to do with the product rule, but why did they take out the 100 like that?
  7. Feb 18, 2008 #6
    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]
  8. Feb 18, 2008 #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?
  9. Feb 18, 2008 #8


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    You really have to know algebra in order to do calculus! 100xy= (100x)y= x(100y)= 100(xy).
  10. Feb 18, 2008 #9
    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. :/
  11. Feb 18, 2008 #10
    Ok, take the derivative of this problem. Treat it as if it's a product.


    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?
  12. Feb 18, 2008 #11
    Ah, of course! Me so dumb. Thanks a lot. :)
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