Derivative of ln x from the limit definition
Let $f(x) = \ln x$. Using the limit definition of the derivative, find $f'(x)$ for general $x>0$, and evaluate $f'(1)$.
Explanation of Method
Rather than simply quoting the known formula $\frac{d}{dx}\ln x = \frac1x$, it can be derived directly from the definition of the derivative as a limit. Forming the difference quotient for $\ln x$ and using the logarithm property $\ln a - \ln b = \ln(a/b)$ combines the two logarithms into one, whose argument is close to $1$ for small $h$. Recognizing the resulting expression as a disguised form of the standard limit $\lim_{u\to0}\frac{\ln(1+u)}{u}=1$ finishes the computation.
- Write down the definition of the derivative for general $x$.
- Combine the two logarithms using $\ln a - \ln b = \ln(a/b)$.
- Rewrite the expression to reveal the standard limit $\lim_{u\to0}\frac{\ln(1+u)}{u}=1$.
- Evaluate $f'(x)$ in general, then substitute $x=1$.
Step 1 · Write down the definition
The derivative of $f(x) = \ln x$ at a general point $x>0$ is:
$$ f'(x) = \lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h}. $$
Step 2 · Combine the logarithms
Using $\ln a - \ln b = \ln\left(\dfrac{a}{b}\right)$:
$$ \ln(x+h) - \ln x = \ln\left(\frac{x+h}{x}\right) = \ln\left(1+\frac{h}{x}\right). $$
So the difference quotient becomes:
$$ f'(x) = \lim_{h\to0} \frac{1}{h}\ln\left(1+\frac{h}{x}\right). $$
Step 3 · Reveal the standard limit
Let $u = \dfrac{h}{x}$, so $h = ux$, and as $h \to 0$ (with $x$ fixed), $u \to 0$ as well. Substituting:
$$ f'(x) = \lim_{u\to0} \frac{1}{ux}\ln(1+u) = \frac{1}{x}\lim_{u\to0}\frac{\ln(1+u)}{u}. $$
The limit $\displaystyle\lim_{u\to0}\frac{\ln(1+u)}{u}=1$ is a standard result (it is, in fact, the derivative of $\ln(1+u)$ at $u=0$, or equivalently follows from $\lim_{u\to0}(1+u)^{1/u}=e$ by taking logarithms).
Step 4 · Evaluate $f'(x)$ and $f'(1)$
Substituting this standard limit:
$$ f'(x) = \frac{1}{x}\cdot 1 = \frac{1}{x}. $$
Evaluating at $x=1$:
$$ f'(1) = \frac{1}{1} = 1. $$
Summary Table
| Quantity | Expression | Value |
|---|---|---|
| Difference quotient | $\dfrac1h\ln\left(1+\dfrac hx\right)$ | — |
| $f'(x)$, general $x>0$ | $\dfrac1x$ | — |
| $f'(1)$ | $\dfrac11$ | $1$ |
Additional Notes
- The substitution $u=\frac hx$ is what converts the difference quotient into exactly the standard limit $\lim_{u\to0}\frac{\ln(1+u)}{u}=1$ — this same limit underlies the derivative of $\ln$ at every point, not just $x=1$.
- This derivation also explains why $e$ is the natural base for logarithms in calculus: the constant $\lim_{u\to0}(1+u)^{1/u}=e$ is precisely what makes the derivative of $\ln x$ equal to the simple expression $\frac1x$, with no extra multiplicative constant.
- Once $f'(x)=\frac1x$ is established for general $x$, evaluating at any specific point (such as $x=1$) is immediate substitution — no further limit computation is needed.
Result
The derivative of $f(x)=\ln x$, and its value at $x=1$, are: