Calculus with Applications (10th Edition)

$$f(x)=\frac{6}{x}$$ Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &=\lim\limits_{h \to 0} \frac{\frac{6}{x+h}-\frac{6}{x}}{h}\\ &=\lim\limits_{h \to 0} \frac{-6h}{xh(x+h)}\\ &=\lim\limits_{h \to 0} \frac{-6}{x(x+h)} \\ &=\frac{-6}{x^{2}} \end{split}$$ Use this result to find $$f^{\prime}(2)=\frac{-6}{(2)^{2}}=\frac{-3}{2}$$ $$f^{\prime}(16)=\frac{-6}{(16)^{2}}=\frac{-6}{4}=\frac{-6}{256}= \frac{-3}{128}$$ and $$f^{\prime}(-3)=\frac{-6}{(-3)^{2}}= \frac{-6}{9}=\frac{-2}{3}$$
$$f(x)=\frac{6}{x}$$ Go through the four steps to find $f^{\prime}(x)$ Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$ $$f(x +h)=\frac{6}{x+h}$$ Step 2 $$f(x +h) -f(x)=\frac{6}{x+h}-\frac{6}{x}$$ Step 3 $$\frac{f(x +h)- f(x)}{h}= \frac{\frac{6}{x+h}-\frac{6}{x}}{h}$$ Step 4 Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &=\lim\limits_{h \to 0} \frac{\frac{6}{x+h}-\frac{6}{x}}{h}\\ &=\lim\limits_{h \to 0} \frac{-6h}{xh(x+h)}\\ &=\lim\limits_{h \to 0} \frac{-6}{x(x+h)} \\ &=\frac{-6}{x^{2}} \end{split}$$ Use this result to find $$f^{\prime}(2)=\frac{-6}{(2)^{2}}=\frac{-3}{2}$$ $$f^{\prime}(16)=\frac{-6}{(16)^{2}}=\frac{-6}{4}=\frac{-6}{256}= \frac{-3}{128}$$ and $$f^{\prime}(-3)=\frac{-6}{(-3)^{2}}= \frac{-6}{9}=\frac{-2}{3}$$