Calculus with Applications (10th Edition)

$$f(x)=\sqrt {x}$$ Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &= \lim\limits_{h \to 0}\frac{\sqrt {x+h} -\sqrt {x}}{h}\\ &=\lim\limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &=\lim\limits_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &=\lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x}) } \\ &=\lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{1}{2\sqrt {x}} \end{split}$$ Use this result to find $$f^{\prime}(2)=\frac{1}{2\sqrt {2}}$$ $$f^{\prime}(16)=\frac{1}{2\sqrt {16}} =\frac{1}{8}$$ and $$f^{\prime}(-3)=\frac{1}{2\sqrt {-3}} \\ \quad\quad\quad\quad[\text { is not a real number , hence }]$$ $f^{\prime}(-3)$ does not exist.
$$f(x)=\sqrt {x}$$ Go through the four steps to find $f^{\prime}(x)$ Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$ $$f(x +h)=\sqrt {x+h}$$ Step 2 $$f(x +h) -f(x)=\sqrt {x+h} -\sqrt {x}$$ Step 3 $$\frac{f(x +h)- f(x)}{h}= \frac{\sqrt {x+h} -\sqrt {x}}{h}$$ Step 4 Now use the rules for limits to get $$\begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &= \lim\limits_{h \to 0}\frac{\sqrt {x+h} -\sqrt {x}}{h}\\ &=\lim\limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &=\lim\limits_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &=\lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x}) } \\ &=\lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{1}{2\sqrt {x}} \end{split}$$ Use this result to find $$f^{\prime}(2)=\frac{1}{2\sqrt {2}}$$ $$f^{\prime}(16)=\frac{1}{2\sqrt {16}} =\frac{1}{8}$$ and $$f^{\prime}(-3)=\frac{1}{2\sqrt {-3}} \\ \quad\quad\quad\quad[\text { is not a real number , hence }]$$ $f^{\prime}(-3)$ does not exist.