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Differentiation of natural logarithms

  1. Oct 24, 2008 #1
    I encountered a proof problem when I was reading up on the derivatives of natural logarithms' section. It gave a rule which said this : [tex]\text{For } a >0 \text{ and } a\ne 1 \text{,}\\\frac{d}{dx}(a^{u}) \ = \ a^{u} \ \ln{a\frac{du}{dx}}[/tex]

    To prove it on my own, I made a few identities:
    [tex]a^{u}=y [/tex]
    [tex] u \ln{a}=\ln{y} [/tex]
    [tex] e^{u\ln{a}}=e^{\ln{y}} \text{ or } e^{u\ln{a}}=y[/tex]

    Now taking the derivative of [tex]\frac{d}{dx}a^{u}[/tex]
    [tex]\frac{d}{dx}a^{u}=\frac{d}{dx}[e^{u\ln{a}}][/tex] I obtained this result by substituting a^u for its identity.
    [tex]\frac{d}{dx}[e^{u\ln{a}}]=(e^{u\ln{a}}) \frac{d[e^{u\ln{a}}]}{dx}[/tex]
    Utilizing the chain rule, I obtained that. But, I do not know how to take the derivative of [tex]e^{u \ln{a}}[/tex] on the right hand side.
    Can anyone give me a simple explanation how to find the derivative of [tex]e^{u \ln{a}}[/tex] ?
  2. jcsd
  3. Oct 24, 2008 #2


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    You are getting the chain rule wrong. d/dx(e^(u*lna))=(e^(u*lna))*d/dx(u*lna). lna is a constant.
  4. Oct 24, 2008 #3
    Sorry, accidentally wrote it wrong.
    \frac{d}{dx}[e^{u\ln{a}}]=(e^{u\ln{a}}) \frac{d}{dx}[u\ln{a}]
    [/tex] is what I meant to write. But, my concern is, how do I simply [tex]\frac{d}{dx}[u\ln{a}][/tex] ?
  5. Oct 24, 2008 #4
    What is ln(a)?
  6. Oct 24, 2008 #5


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    ln a is a constant. Just pull it out. d/dx(u*ln(a))=ln(a)*du/dx.
  7. Oct 24, 2008 #6
    A constant. I get [tex](e^{u\ln{a}}) \frac{d[u\ln{a}]}{dx}= (e^{u\ln{a}}*\ln{a}) \frac{du}{dx}= (a^{u}*\ln{a}) \frac{du}{dx}[/tex] which exactly matches the rule. Thanks.
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