Find y' for y = e^x

Find $y'$ for $y = e^{x}$.

Step 1: Write the difference quotient

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

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

Using the law of exponents $e^{x+\Delta x} = e^{x}e^{\Delta x}$: \[\frac{\Delta y}{\Delta x} = \frac{e^{x}\bigl(e^{\Delta x}-1\bigr)}{\Delta x} = e^{x}\cdot\frac{e^{\Delta x}-1}{\Delta x}.\]

The factor $e^{x}$ does not depend on $\Delta x$, so it may be taken outside the limit.

Step 3: Take the limit

Using the standard limit $\displaystyle\lim_{u\to 0}\frac{e^{u}-1}{u} = 1$: \[y' = \lim_{\Delta x \to 0} e^{x}\cdot\frac{e^{\Delta x}-1}{\Delta x} = e^{x}\cdot 1 = e^{x}.\]

Step 4: Verification

\[x = 0:\quad y'(0) = e^{0} = 1, \text{ in agreement with Problem 7} \;\;\checkmark\]

\[x = 1,\ \Delta x = 10^{-6}:\quad \frac{e^{1+10^{-6}}-e}{10^{-6}} \approx 2.71828 \approx e \;\;\checkmark\]