Question

ex 3 if f(x) = √x, find f(x) and state its domain f(x + h)= √x + h

Explanation:

Step1: Use the definition of the derivative

The definition of the derivative $f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\sqrt{x}$ and $f(x + h)=\sqrt{x + h}$, we have $f'(x)=\lim_{h
ightarrow0}\frac{\sqrt{x + h}-\sqrt{x}}{h}$.

Step2: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{x + h}+\sqrt{x}$. So we get $f'(x)=\lim_{h
ightarrow0}\frac{(\sqrt{x + h}-\sqrt{x})(\sqrt{x + h}+\sqrt{x})}{h(\sqrt{x + h}+\sqrt{x})}$. Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(x + h)-x=h$. Then $f'(x)=\lim_{h
ightarrow0}\frac{h}{h(\sqrt{x + h}+\sqrt{x})}=\lim_{h
ightarrow0}\frac{1}{\sqrt{x + h}+\sqrt{x}}$.

Step3: Evaluate the limit

As $h
ightarrow0$, we substitute $h = 0$ into the expression. So $f'(x)=\frac{1}{2\sqrt{x}}$.

Step4: Find the domain of $f'(x)$

The function $y = f'(x)=\frac{1}{2\sqrt{x}}$ is undefined when $x<0$ (because of the square - root in the denominator) and when $x = 0$ (since $\frac{1}{2\sqrt{0}}$ is undefined). So the domain of $f'(x)$ is $(0,+\infty)$.

Answer:

$f'(x)=\frac{1}{2\sqrt{x}}$, domain: $(0,+\infty)$