Question

Use the definition of the derivative to find $f\text{'}$for each function $f$in Exercises 34-59

$f\left(x\right)=3\sqrt{x}$

Short answer

The value of$f\text{'}\left(x\right)=\frac{3}{2\sqrt{x}}$

Step-by-step solution

Step 1. Given information

The given function$f\left(x\right)=3\sqrt{x}$

Step 2. Finding the value of f'(x)

We know that $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}.........\left(1\right)$

Given $f\left(x\right)=3\sqrt{x}$

Then $f(x+h)=3\sqrt{x+h}$

putting these values in (1)

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{3\sqrt{x+h}-3\sqrt{x}}{h} \\ =\underset{h\to 0}{\lim }\frac{3\sqrt{x+h}-3\sqrt{x}}{h}\times \frac{3\sqrt{x+h}+3\sqrt{x}}{3\sqrt{x+h}+3\sqrt{x}}\left[Rationalizing\right] \\ =\underset{h\to 0}{\lim }\frac{{\left(3\sqrt{x+h}\right)}^2-{\left(3\sqrt{x}\right)}^2}{h\left[3\right(\sqrt{x+h}+\sqrt{x}\left)\right]}[\because (a+b\left)\right(a-b)=a^2-b^2] \\ =\underset{h\to 0}{\lim }\frac{9(x+h)-9x}{h\left[3\right(\sqrt{x+h}+\sqrt{x}\left)\right]} \\ =\underset{h\to 0}{\lim }\frac{9x+9h-9x}{h\left[3\right(\sqrt{x+h}+\sqrt{x}\left)\right]} \\ =\underset{h\to 0}{\lim }\frac{9h}{3h\left[\right(\sqrt{x+h}+\sqrt{x}\left)\right]} \\ =\underset{h\to 0}{\lim }\frac{3}{(\sqrt{x+h}+\sqrt{x})} \end{gathered}$$

Putting $h=0$

$$\begin{gathered} =\frac{3}{\sqrt{x+0}+\sqrt{x}} \\ =\frac{3}{\sqrt{x}+\sqrt{x}} \\ =\frac{3}{2\sqrt{x}} \end{gathered}$$

Hence,the value of$f\text{'}\left(x\right)=\frac{3}{2\sqrt{x}}$