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Hyperbolic Differentiation

  1. Oct 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Differentiate gif.gif

    2. Relevant equations
    Chain Rule: dg/dx = du/dx . dv/du . dg/dv

    3. The attempt at a solution
    My answer(wrong):
    e33ab0e3d5.jpg

    Correct answer provided to us(not mine):
    2be4cb75b9.png

    I understand the correct solution that was provided to us, but what I don't understand is why my method isnt correct? Also can you check in particular my dv/du . I suspect theres something wrong there.
     
  2. jcsd
  3. Oct 27, 2015 #2

    pasmith

    User Avatar
    Homework Helper

    The derivative of [itex]3^v = e^{v \log 3}[/itex] is [itex](\log 3) e^{v \log 3} = (\log 3)3^v[/itex], not [itex]3v^{v-1}[/itex] which appears to be what you have.

    If you set [itex]v = u\log x[/itex] then you want to calculate [tex]\frac{dv}{du} = \log x + \frac ux \frac{dx}{du}[/tex] and the [itex]\frac{dx}{du}[/itex] cancels with [itex]\frac{du}{dx}[/itex] when you apply the chain rule.

    Differentiating functions in exponents is straightforward: By the chain rule, with [itex]a > 0[/itex], [tex]
    \frac{d}{dx} a^{f(x)} = (\log a)a^{f(x)} f'(x).[/tex]
     
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