Find y' for y = e^-x

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

Step 1: Write the difference quotient

Using $e^{-(x+\Delta x)} = e^{-x}e^{-\Delta x}$ and factoring out $e^{-x}$: \[\frac{\Delta y}{\Delta x} = \frac{e^{-(x+\Delta x)} - e^{-x}}{\Delta x} = \frac{e^{-x}e^{-\Delta x} - e^{-x}}{\Delta x} = e^{-x}\cdot\frac{e^{-\Delta x}-1}{\Delta x}.\]

Step 2: Evaluate the limit

Substitute $u = -\Delta x$, which tends to $0$ together with $\Delta x$: \[\frac{e^{-\Delta x}-1}{\Delta x} = \frac{e^{u}-1}{-u} = -\frac{e^{u}-1}{u} \;\xrightarrow[u\to 0]{}\; -1,\]

using the standard limit $\displaystyle\lim_{u\to 0}\frac{e^{u}-1}{u} = 1$. Therefore

\[ y' = e^{-x}\cdot(-1) = -e^{-x}. \]

Step 3: Verification

The chain rule with $u = -x$ confirms the answer: \[y = e^{u},\quad u = -x \;\Longrightarrow\; y' = e^{u}\cdot u' = e^{-x}\cdot(-1) = -e^{-x} \;\;\checkmark\]

\[x = 0,\ \Delta x = 10^{-6}:\quad \frac{e^{-10^{-6}}-1}{10^{-6}} \approx -1.0 \;\;\checkmark\]