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Chapter 6: Problem 2

Find the derivative of \(y=f(x)=\log _{e} x\), using first principle.

Short Answer

Expert verified
The derivative of \(y = \log_e x\) is \(\frac{1}{x}\)

Step by step solution

01

Write down the function and it's increment

We start with the function \(y = f(x) = \log_e x\). We write its increment as \(f(x + \Delta x) = \log_e (x + \Delta x)\).
02

Write down the difference quotient

Next, we write down the difference quotient which is \(\frac{f(x + \Delta x) - f(x)}{\Delta x}\). Replacing \(f(x + \Delta x)\) and \(f(x)\) by their respective expressions, we get \(\frac{\log_e (x + \Delta x) - \log_e x}{\Delta x}\).
03

Use logarithm properties

We use the property of logarithms, \( \log_b(m) - \log_b(n) = \log_b(\frac{m}{n}) \), to simplify the difference quotient to \(\frac{\log_e(\frac{x + \Delta x}{x})}{\Delta x} = \frac{\log_e(1+\frac{\Delta x}{x})}{\Delta x}\).
04

Change of Variable

Let \(\Delta x = ux\). Then our expression becomes \(\frac{\log_e(1+u)}{ux}\). Pythagoras theorem tells us that the derivative of \(y = \log_e x\) will exist if and only if the limit as \( u\rightarrow 0\) of the above expression exists. This gives us a new limit problem: find \(\lim_{u\rightarrow 0}[\frac{\log_e(1+u)}{ux}]\).
05

Evaluate the limit using l'hopital's rule

To evaluate the limit, we need to see if we can apply l'Hopital's rule. We can, since the limit of the form is \(0/0\). L'Hopital's rule states that \(\lim_{x\rightarrow a}[\frac{f(x)}{g(x)}] = \lim_{x\rightarrow a}[\frac{f'(x)}{g'(x)}]\). After applying l'hopital's rule to our limit, we get \(\lim_{u\rightarrow 0}[\frac{1/(1+u)}{x}] = \lim_{u\rightarrow 0}[\frac{1}{x(1+u)}] = \frac{1}{x} = \frac{1}{x}\) as the derivative of \(y = \log_e x\).

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