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Derivative e^(x/2)^2

  1. Dec 18, 2008 #1
    Hi

    Can you please help me with this derivative:

    e ^ (x/2) ^2


    I will have an exame in 3 hours and i want to know how to resolve this derivative.


    Thanks to all
     
  2. jcsd
  3. Dec 18, 2008 #2
    You need to apply the chain rule three times.
     
  4. Dec 18, 2008 #3
    It's ambiguious how you have it written. I assume you mean...

    e ^ ((x/2) ^2)

    .Your Outside Function is..

    e^(....)

    The function inside that one is...

    (.......)^2
     
  5. Dec 18, 2008 #4
    yes, that's right...
    How it can be resolved?
     
  6. Dec 18, 2008 #5

    D H

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    That would be the standard interpretation. Without parentheses exponentiation evaluates right-to-left.[/QUOTE]
    Only one application of the chain rule is needed (if the expression to be differentiated is [itex]\exp((x/2)^2)[/itex], that is). Do you really need the chain rule to compute the derivative of (x/2)^2=x^2/4?
     
  7. Dec 18, 2008 #6
    Remember (and prove for yourself) that if

    [tex] f(x) = e^{g(x)} [/tex]

    Then,

    [tex] \frac{df}{dx} = e^{g(x)} \frac{dg}{dx} [/tex]
     
  8. Dec 18, 2008 #7


    So the derivative of (x/2)^2 is x?
     
  9. Dec 18, 2008 #8
    Pretty unlikely. Remember that someone already told you that if you did not want to use the chain rule to rewrite

    [tex] \left(\frac{x}{2}\right)^{2} = \frac{x^2}{4} [/tex]

    If you do want to use the chain rule then remember that you would get

    [tex] \left[\left(\frac{x}{2}\right)^{2}\right]^{'} = 2 \cdot \left(\frac{x}{2}\right) \cdot \frac{1}{2} [/tex]

    Since you have to differentiate "the inside" as well. That's what the chain rule is all about :)
     
  10. Dec 18, 2008 #9
    Ok, thanks.
    You have help me a lot...
    You're great..

    Thanks
     
  11. Dec 21, 2008 #10
    [tex]\frac{df}{dx} = e^{g(x)} \frac{dg}{dx} [/tex]
    That holds if it s e^(x). The more general form is slightly different:

    f(x) = n^h(x) where n is any number
    f'(x)= n^h(x) * ln(n) * h'(x)

    when n = e, ln(n) = 1.

    Even more generally, the derivatve of f(x) = r(x)^h(x) can be shown easily by the following proof:
    f(x) = r(x)^h(x)
    ln(f(x)) = ln(r(x)^h(x))
    ln(f(x)) = h(x) ln(r(x))
    f'(x)/f(x) = h(x)r'(x)/r(x) + h'(x)ln(r(x))
    f'(x) = f(x) [h(x)r'(x)/r(x) + h'(x)ln(r(x))]

    f'(x) = r(x)^h(x) [h(x)r'(x)/r(x) + h'(x)ln(r(x))]
     
    Last edited: Dec 21, 2008
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