Differentiation of natural logarithms

[razored]

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

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

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

 

[Dick]

You are getting the chain rule wrong. d/dx(e^(u*lna))=(e^(u*lna))*d/dx(u*lna). lna is a constant.

 

[razored]

Sorry, accidentally wrote it wrong.
$$\frac{d}{dx}[e^{u\ln{a}}]=(e^{u\ln{a}}) \frac{d}{dx}[u\ln{a}]$$ is what I meant to write. But, my concern is, how do I simply $$\frac{d}{dx}[u\ln{a}]$$ ?

 

[Tedjn]

What is ln(a)?

 

[Dick]

ln a is a constant. Just pull it out. d/dx(u*ln(a))=ln(a)*du/dx.

 

[razored]

A constant. I get $$(e^{u\ln{a}}) \frac{d[u\ln{a}]}{dx}= (e^{u\ln{a}}*\ln{a}) \frac{du}{dx}= (a^{u}*\ln{a}) \frac{du}{dx}$$ which exactly matches the rule. Thanks.