Question

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

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

Short answer

The value of$f\text{'}\left(x\right)=\frac{-9}{6x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Step-by-step solution

Step 1. Given information

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

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

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

Putting these values in (1)

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

Putting $h=0$

$$\begin{gathered} =\frac{-9}{\sqrt{x}\sqrt{x+0}(3\sqrt{x}+3\sqrt{x+0})} \\ =\frac{-9}{\sqrt{x}\sqrt{x}(3\sqrt{x}+3\sqrt{x})} \\ =\frac{-9}{x(2.3\sqrt{x})} \\ =\frac{-9}{6x^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

Hence,the value of$f\text{'}\left(x\right)=\frac{-9}{6x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$