91Ó°ÊÓ

The definition of the derivative of a function \(f(x)\) is: $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ Our function is \(f(x) = e^{-x}\). We must compute \(f(x+h)\) and insert it into the formula: $$f(x + h) = e^{-(x + h)} = e^{-x}e^{-h}$$ Now, substitute the function expressions into the derivative definition: $$f'(x) = \lim_{h \to 0} \frac{e^{-x}e^{-h} - e^{-x}}{h}$$

In the limit expression, factor out the common factor \(e^{-x}\): $$f'(x) = e^{-x} \lim_{h \to 0} \frac{e^{-h} - 1}{h}$$

We are given a hint to use the fact that: $$\lim_{h \to 0} \frac{e^{h} - 1}{h} = 1$$ Now recognize that the given limit can be rewritten using this fact by noting that \(e^{-h} = \frac{1}{e^h}\): $$\lim_{h \to 0} \frac{e^{-h} - 1}{h} = \lim_{h \to 0} \frac{\frac{1}{e^h} - 1}{h} = \lim_{h \to 0} \frac{1 - e^h}{h\cdot e^h}$$ Multiply both the numerator and denominator by \(-1\) so we can use the given fact: $$\lim_{h \to 0} \frac{1 - e^h}{h\cdot e^h} = \lim_{h \to 0} \frac{e^h - 1}{ (-h)\cdot e^h}$$ Now, use the fact that the limit of products is the product of limits and that \(e^x\) is continuous for all \(x\): $$\lim_{h \to 0} \frac{e^h - 1}{ (-h)\cdot e^h} = -\lim_{h \to 0} \frac{e^h - 1}{h} \cdot \lim_{h \to 0} e^h = -1 \cdot 1 = -1$$

Now, substitute the value of the limit back into the expression for \(f'(x)\) from Step 2: $$f'(x) = e^{-x} \cdot (-1)$$ So the derivative of the function \(f(x) = e^{-x}\) is: $$f'(x) = -e^{-x}$$