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How do i differentiate something like 12/(x^+3)?2

  1. Dec 5, 2011 #1
    How do i differentiate something like 12/(x^+3)??2

    I am acquainted with differentiating equations like (2x+5)^2 ( by chain rule and the direct method). But how do i go about with the equation 12/(x^2+3)? The specific doubt is that the answer given -12(x^2+3)^-2 into(multiplied by) 2x. Im getting the answer, but its not multiplied by 2x, so i get -12(x^2+3)^-2. I dont know why
     
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  3. Dec 6, 2011 #2

    dextercioby

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    Re: How do i differentiate something like 12/(x^+3)??2

    Do you know about the chain rule ? What does it say ?
     
  4. Dec 6, 2011 #3
    Re: How do i differentiate something like 12/(x^+3)??2

    the chain rule is a formula for computing the derivative of the composition of two or more functions. But i dont want the chain rule method(if there is one in this case) for getting the answer, i want the direct method, or is chain rule the only way of doin it?
     
  5. Dec 6, 2011 #4
    Re: How do i differentiate something like 12/(x^+3)??2

    hi mutineer -

    not sure what you mean by "direct method", but consider re-writing the expression as:

    [itex]f(x) = \frac{12}{(x^2 + 3)} = 12\cdot (x^2 + 3)^{-1}[/itex]

    If you define the denominator [itex] (x^2 + 3) := g(x)[/itex], then f(x) becomes:

    [itex] f(x) = 12\cdot \frac{1}{g(x)} = 12*\cdot g(x)^{-1}[/itex]

    (I wrote this both as a fraction and using negative exponents to avoid confusion with the inverse f'n).

    In your "direct method", you've computed the first part of the chain rule, i.e. you've found:

    [itex] = \frac{d}{dx}f(x) = 12 \cdot \frac{d}{dx}g(x)^{-1} = 12 \cdot (-1)\cdot g(x)^{-2}[/itex]

    But that /isn't/ the full derivative. Recall the chain rule requires you to multiply by the inside derivative as well. Here, the inside derivative would be g'(x) (the derivative of g(x) w.r.t. x; which I'll leave to you to sort out).

    Hope this helps!
     
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