First Principals Logarithmic Derivative

[ChaoticLlama]

We did derivatives of logarithmic functions just recently, and of course we were given the first principals definition. However, there was one line which was unsatisfying. He said to 'just trust him'. He was aware at the time that he it was a step that would leave the class in the dark, but was necessary to complete the definition.

Anyway, if it is not way beyond my mathematical knowledge at this point, could you please explain the indicated line.

By Definition:
\(\displaystyle \frac{{df}}{{dx}} = {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}\)

Here, \(\displaystyle f(x) = \log _a (x)\)

\(\displaystyle \begin{array}{c}
\frac{{df}}{{dx}} = {\lim }\limits_{h \to 0} \frac{{\log (x + h) - \log (x)}}{h} \\
={\lim }\limits_{h \to 0} \frac{{\log (\frac{{x + h}}{x})}}{h} \\
= {\lim }\limits_{h \to 0} \frac{{\log (1 + \frac{h}{x})}}{h} \\
\end{array}\)

Let z = h/x --> h = zx

\(\displaystyle \begin{array}{c}
\frac{{df}}{{dx}} ={\lim }\limits_{z \to 0} \frac{{\log _a (1 + z)}}{{zx}} \\
= \frac{1}{x}{\lim }\limits_{z \to 0} \log _a (1 + z)^{\frac{1}{z}} \\
= \frac{1}{x} {\lim }\limits_{z \to 0} \log _a \left[ {{\lim }\limits_{z \to 0} (1 + z)^{\frac{1}{z}} } \right]*** \\
= \frac{1}{x}\log _a e \\
\end{array}\)

The Line with the 3 ***, where he put the limit inside the arguement. He told us that it was allowable 'because the function is continuous and differentiable at all values of x > 0' and to trust him

If it is possible to explain why such an operation is allowable/works would be appreciated .

 

[soroban]

Hello, ChaoticLlama!

The simplest explnanation is that your teacher used a defintion.

One of the definitions of the transcendental number \(\displaystyle e\) is:

\(\displaystyle \;\;\;\lim_{z\to0}\left(1\,+\,z\right)^{\frac{1}{z}}\;=\;2.718281828459045...\;=\;e\)

Another is: \(\displaystyle \:\lim_{x\to\infty}\left(1\,+\,\frac{1}{x}\right)^x\)

 

[ChaoticLlama]

Sorry if I was a bit unclear in my question.

I do know the limit of e.

I was curious under what conditions you are allowed to put a limit in front of the arguement and how/why that works in solving the problem.

For my understanding, log(x), is similar to sin(x) and sqrt(x) where an operation is performed on a number.
Yet in this case, the number has been 'interrupted' from being evaluated by the log function.

Is this explanation something I will come across in higher math education? Or is it explainable but it was only left out due to time constraints?

Thank you for your input.

 

[pka]

I think that I understand your question.
Here is a theorem: If f is continuous at b and \(\displaystyle \ \rm{lim}\limits_{x \to a} g(x) = b\) then \(\displaystyle \ \rm{lim}\limits_{x \to a} f\left( {g(x)} \right) = f(b)\).

In other words: \(\displaystyle \ {{\rm{lim}}}\limits_{x \to a} f\left( {g(x)} \right) = f\left( {\ {{\rm{lim}}}\limits_{x \to a} g(x)} \right)\)

 

[ChaoticLlama]

Thanks, that clears it up a bit.