Why is the derivative of ln x 1/x?

Steps to Solve

1. Recall the definition of the derivative

The derivative of a function $f(x)$ can be defined as the limit of the difference quotient:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

We will apply this definition to the function $f(x) = \ln x$.

2. Apply the definition of the derivative to ln x

Substituting $f(x) = \ln x$ into the derivative formula, we have:

$$ f'(x) = \lim_{h \to 0} \frac{\ln(x + h) - \ln x}{h} $$

3. Use the property of logarithms to simplify

Using the property of logarithms that states $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$, we rewrite the equation:

$$ f'(x) = \lim_{h \to 0} \frac{\ln\left(\frac{x + h}{x}\right)}{h} $$

4. Simplify the argument of the logarithm

Now we simplify the argument of the logarithm:

$$ \frac{x + h}{x} = 1 + \frac{h}{x} $$

Thus, we have:

$$ f'(x) = \lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x}\right)}{h} $$

5. Use the limit property of ln(1 + u)

As $h \to 0$, we can set $u = \frac{h}{x}$ which implies that as $h \to 0$, $u \to 0$. Therefore, we modify our limit:

$$ f'(x) = \lim_{u \to 0} \frac{\ln(1 + u)}{xu} $$

6. Apply the known limit of ln(1 + u)

We know that:

$$ \lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1 $$

Substituting back gives us:

$$ f'(x) = \frac{1}{x} \cdot 1 = \frac{1}{x} $$