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Applying the Chain Rule

  1. Oct 30, 2005 #1
    I've been having some trouble grasping the conditions necessary to apply the chain rule to achieve the derivative of an algebraic expression or even apply it to a real world situation.

    So, my question to those skilled in qualitatively explaining the conditions for applying the Chain Rule and also when the Product Rule or Quotient Rule should be applied when the Chain Rule won't work.

    Essentially, I believe the Chain Rule is applied when it is possible to seperate a function into two seperate algebraic equations. Is this some sort of shortcut to not using the Product Rule or Quotient Rule in order to obtain the derivative of an equation? Is there more to that definition than I suspect?

    Are there situations where the Chain Rule cannot be used to obtain the derivative of a function?

    Thanks in advance.
     
  2. jcsd
  3. Oct 30, 2005 #2
    I'm sure someone here has a better explanation, but this is how I understand it: The chain rule is generally used for composite functions, for example:

    [tex]\sin{e^{2\pi x}}[/tex]

    ...is really [itex]f\left(g\left(x\right)\right)[/itex] where:

    [tex]f\left(x\right)=\sin{x};\quad g\left(x\right)=e^{2\pi x}[/tex]

    Whereas the product and quotient rules would be used for something like:

    [tex]\frac{\sin{x}}{e^{2\pi x}}[/tex]
     
  4. Oct 30, 2005 #3
    therrem: Chain Rule

    if g is differentiable at point x and f is differentiable at the point g(x), then the composition of f*g is differentiable at the point x. Moreover, if

    y = f(g(x)) qnd u = g(x)

    then y = f(u)

    and

    dy/dx = dy/du * du/dy
     
  5. Oct 30, 2005 #4
    Chain Rule works indeed for what seems to be a composite function or something that is made out of more components. For example a trigonometric function.

    To find the first derivative of y=tan(2x+x^2) for example you must apply the Chain Rule to get a correct answer.

    dy/dx=Sec^2(2x+x^2)*(2x+2)
     
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