## Calculus with Applications (10th Edition)

$$f(x)=e^{x}$$ Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &=\lim\limits_{h \to 0} \frac{e^{x+h}-e^{x} }{h}\\ &=\lim\limits_{h \to 0} \frac{e^{x}(e^{h}-1) }{h}\\ & \quad\quad\quad\quad[\text { by L'Hospital's Rule }] \\ &=e^{x} \end{split}$$ Use this result to find $$f^{\prime}(2)=e^{2}=7.3891;$$ $$f^{\prime}(16)=e^{16}= 8,886,111;$$ and $$f^{\prime}(-3)=e^{-3}= 0.0498.$$
$$f(x)=e^{x}$$ Go through the four steps to find $f^{\prime}(x)$ Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$ $$f(x +h)=e^{x+h}$$ Step 2 $$f(x +h) -f(x)=e^{x+h}-e^{x}$$ Step 3 $$\frac{f(x +h)- f(x)}{h}= \frac{e^{x+h}-e^{x} }{h}$$ Step 4 Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &=\lim\limits_{h \to 0} \frac{e^{x+h}-e^{x} }{h}\\ &=\lim\limits_{h \to 0} \frac{e^{x}(e^{h}-1) }{h}\\ & \quad\quad\quad\quad[\text { by L'Hospital's Rule }] \\ &=e^{x} \end{split}$$ Use this result to find $$f^{\prime}(2)=e^{2}=7.3891,$$ $$f^{\prime}(16)=e^{16}= 8,886,111$$ and $$f^{\prime}(-3)=e^{-3}= 0.0498$$