Find y' for y = log_a x, where a > 0, a ≠ 1, x > 0

Find $y'$ for $y = \log_{a} x$, where $a \gt 0$, $a \neq 1$, $x \gt 0$.

Step 1: Write the difference quotient

Using the law $\log_a A - \log_a B = \log_a\dfrac{A}{B}$: \[\frac{\Delta y}{\Delta x} = \frac{\log_{a}(x+\Delta x) - \log_{a} x}{\Delta x} = \frac{1}{\Delta x}\log_{a}\!\left(\frac{x+\Delta x}{x}\right) = \frac{1}{\Delta x}\log_{a}\!\left(1+\frac{\Delta x}{x}\right).\]

Step 2: Introduce a substitution

Let $\alpha = \dfrac{\Delta x}{x}$, so that $\Delta x = \alpha x$ and $\alpha \to 0$ as $\Delta x \to 0$. Then

\[ \frac{\Delta y}{\Delta x} = \frac{1}{\alpha x}\log_{a}(1+\alpha) = \frac{1}{x}\cdot\frac{1}{\alpha}\log_{a}(1+\alpha) = \frac{1}{x}\log_{a}(1+\alpha)^{1/\alpha}. \]

Step 3: Take the limit

Using the fundamental limit $\displaystyle\lim_{\alpha\to 0}(1+\alpha)^{1/\alpha} = e$ and the continuity of $\log_{a}$: $y' = \lim_{\alpha\to 0}\frac{1}{x}\log_{a}(1+\alpha)^{1/\alpha} = \frac{1}{x}\log_{a} e$.

Step 4: Simplify by change of base

Since $\log_{a} e = \dfrac{1}{\ln a}$: $y' = \frac{1}{x}\cdot\frac{1}{\ln a} = \frac{1}{x\ln a}$.

Step 5: Verification

\[a = e:\quad y' = \frac{1}{x\ln e} = \frac{1}{x} = (\ln x)' \;\;\checkmark\]

\[\text{Inverse of Problem 18:}\quad x = a^{y} \;\Longrightarrow\; 1 = a^{y}\ln a\cdot\frac{dy}{dx} \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{x\ln a} \;\;\checkmark\]