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Finding Derivatives

  1. Jan 3, 2006 #1
    Find dy/dx if y = 4^(-x + 3)

    I really don't know how to go about doing this.. never had to take the derivative of something with a variable as an exponent..

    Thanks for the help in advance.
  2. jcsd
  3. Jan 3, 2006 #2


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    This derivative requires the logarithmic change of base rule.
    log_ax = lnx/lna
    So this would become:
    F(x) = 4^(3-x)
    F'(x) = ln4 * (-4^(3-x))
    = -2ln2 * 4^(3-x)
  4. Jan 3, 2006 #3


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    The derivative of [itex]a^x[/itex] is [itex]\ln{a} \cdot a^x[/itex] but don't forget the chain rule.
    Last edited: Jan 3, 2006
  5. Jan 3, 2006 #4
    You probably meant that the derivative of [itex]a^x[/itex] is [itex]\ln{a} \cdot a^x[/itex]?
  6. Jan 3, 2006 #5


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    Of course (it's still morning here :blushing:)

  7. Jan 3, 2006 #6


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    Or just: if y= ax, then ln(y)= x ln(a). Differentiate both sides, using the chain rule on the left.
  8. Jan 3, 2006 #7

    ok i understand that the formula for derivative of a^x is ln a * a^x

    so i understand the step
    F'(x) = ln 4 * (-4^(3-x)) <--- how did the 4 become negative?
    and then i don't understand the next step at all...
  9. Jan 3, 2006 #8


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    The 4 becomes -4 because of the chain rule, since it says (3-x) and not just x, you have to multiply with the derivative of (3-x), which is -1 causing the sign change.
    Then he just used a property of logaritms, being that ln(a^b) = b.ln(a) so ln(4) = ln(2²) = 2ln(2).
  10. Jan 3, 2006 #9
    ah, thanks a lot :)
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