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

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

Step 1: Write the difference quotient

\[ \frac{\Delta y}{\Delta x} = \frac{a^{x+\Delta x} - a^{x}}{\Delta x}. \]

Step 2: Factor out $a^{x}$

\[ \frac{\Delta y}{\Delta x} = a^{x}\cdot\frac{a^{\Delta x}-1}{\Delta x}. \]

Step 3: Evaluate the limit

Write $a^{\Delta x} = e^{\Delta x\ln a}$ and substitute $u = \Delta x\ln a$, which tends to $0$ together with $\Delta x$: \[\frac{a^{\Delta x}-1}{\Delta x} = \frac{e^{\Delta x\ln a}-1}{\Delta x} = \ln a\cdot\frac{e^{u}-1}{u} \;\xrightarrow[u\to 0]{}\; \ln a\cdot 1 = \ln a.\]

Therefore $y' = \lim_{\Delta x \to 0} a^{x}\cdot\frac{a^{\Delta x}-1}{\Delta x} = a^{x}\ln a$.

Step 4: Verification

\[a = e:\quad y' = e^{x}\ln e = e^{x}\cdot 1 = e^{x}, \text{ agreeing with Problem 17} \;\;\checkmark\]

\[a = 2,\ x = 0,\ \Delta x = 10^{-6}:\quad \frac{2^{10^{-6}}-1}{10^{-6}} \approx 0.69315 \approx \ln 2 \;\;\checkmark\]