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Implicit Differentiation is so confusing

  1. Oct 5, 2007 #1
    Hello there! Please help me relieve my confusion. Thanks!

    For [tex] \frac{d}{dx}[y^{3}] [/tex], why do you need to use the chain rule on this equation? Basically, the chain rule is used on almost every function right? It is just that we do not see the dx/dx since it equals one, for example:

    [tex] f(x)=\sin{x}[/tex] In full notation, its derivative is suppose to look like this right?:
    [tex] f'(x)=(\cos{x}) \frac{dx}{dx}[/tex]. Then the derivative of x with respect to x is just 1.
     
  2. jcsd
  3. Oct 5, 2007 #2

    mjsd

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    sure, if you want to look at it that way.

    nothing wrong with being "explicit".... but after while it would become a nuisance too :smile:

    for implicit diff... all you are doing is to write thing using symbols (for you don't know yet what the function is exactly)...but since from chain rule you know what it form should be so you can write: eg.

    if [tex]y=y(x), [/tex] then
    [tex]\frac{d}{dx}(x^2 y) = 2xy + x^2 \frac{d}{dx} (y) = 2xy +x^2\frac{dy}{dx}[/tex]

    so until u know what y(x) really is you can't do that dy/dx part and you can only leave it as that form until more info is given.
     
  4. Oct 6, 2007 #3
    So really, when you find the derivative, even it is a simple equation like f(x)=x^3, you really are using the chain rule (except that teachers/authors just simplify it to the Power Rule since x does not have a coefficient)??

    Is there logic behind all of these rules and theorems or do i just have to remember it? =[
     
  5. Oct 6, 2007 #4

    Gib Z

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    Well some sort of an understanding for the chain rule is, all you are doing is multiplying the composing rates of change to get the total rate. For example, is I am going twice as fast as You are, and msjd is going 3 times as fast as me, how fast is msjd going compared to you? 6 times, because we multiply both rates. Thats really all the chain rule does.
     
  6. Oct 6, 2007 #5

    Hurkyl

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    What precisely do you mean? Surely you've seen the proofs of these rules and theorems, right?

    But the point is that the derivative laws are the tools to working efficiently with derivatives. It's much like how you learned the laws of arithmetic when you were a little kid, so that you could manipulate numbers efficiently.
     
  7. Oct 6, 2007 #6
    Yes, there are proofs in my text but I do not really bother with them. Should I try to understand and actually attempt to prove it? Will that help my comprehension of their nature?

    Let me elaborate on my last example a bit: If f(x)=x^3, then to find the derivative of f you need to use the "chain rule" right? Basically you need to use the chain rule if you can see any type of composition (i.e. g(x)=x^3, h(x)=x, f(x)=g(h(x))). Is that why my teacher tells us that [tex]\frac{d}{dx} [x^3]= 3x^{2} * \frac{dx}{dx}[/tex], where dx\dx =1?? Isn't that basically the chain rule? So do you always apply the chain rule where there is a composite function? (except I know we can simplify for derivatives of x with no coefficient because dx/dx=1)

    Am I thinking right so far? Thanks for your guys's help!
     
  8. Oct 6, 2007 #7

    CompuChip

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    You could work through the proofs once if you're interested in seeing them. But actually, nobody ever bothers to derive [itex]x \mapsto x^3[/itex] by the chain rule, but just uses [itex]\frac{d}{dx} x^n = n x^{n-1}[/itex]. These are simple rules that make life easier, so you should just remember them. The point is that if you forget them, you can always re-derive them from the chain rule.

    In principle you are right about the dx/dx thing as well. In fact, you can take it as far as you like.
    Define [itex]x_i(x) = x[/itex] for i = 1, 2, 3, ... n up to any n you want. Then if g(x) = x3 I can always write it as
    [tex]f(x) = g(x_1(x_2(\cdots(x_n(x))\cdots))[/tex].
    Taking the derivative would get me
    [tex]\frac{d}{dx} f(x) = \frac{d g(x)}{dx} \underbrace{\frac{dx_1(x)}{dx}}_{=1} \underbrace{\frac{dx_2(x)}{dx}}_{=1} \cdots \underbrace{\frac{dx_n(x)}{dx}}_{=1} = \frac{d g(x)}{dx}. [/tex]

    I could even take [itex]x_i = a x + b, x_j = (x - b)/a [/itex] for some indices i, j and some non-zero number a and any number b. Then since [itex]x_i(x_j(x)) = x_i(a x - b) = ((a x - b) - b) / a = x[/itex] nothing has changed, and you can show that one of the derivatives from the chain rule yields a factor a while another one gives 1/a cancelling them out. But this would be a rather pointless exercise, I must admit. :smile:
     
  9. Oct 6, 2007 #8

    mjsd

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    if you are bored... you can always try:

    another chain rule example:

    [tex]f(x) = x^3[/tex] and let [tex]u = x^3[/tex] then
    [tex]\frac{d}{dx}(f(x)) = \frac{df}{du}\frac{du}{dx} = 1\times 3x^2 =3x^2[/tex]

    or even
    [tex]v = x^2 \Rightarrow f(x) = v^{3/2}[/tex] then
    [tex]\frac{d}{dx}(f(x)) = \frac{df}{dv}\frac{dv}{dx} =
    \frac{3}{2} v^{1/2} \times 2x = 3(x^2)^{1/2} x = 3x^2[/tex]

    :smile:
     
  10. Oct 6, 2007 #9

    Hurkyl

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    Actually, I have seen tables of derivative rules that use the chain rule in every rule. For example, you would not see
    [tex]\frac{d}{dx} x^n = n x^{n-1}[/tex]
    in the table; you would see the power rule written as
    [tex]\frac{d}{dx} u^n = n u^{n-1} \frac{du}{dx}.[/tex]
     
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