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](adsbygoogle = window.adsbygoogle || []).push({});

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] ?

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# Differentiation of natural logarithms

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