Question

Find the derivatives of the function: $f\left(x\right)=\frac{1-{\displaystyle \frac{1}{x}}}{\sqrt{x}}.$

Short answer

The derivative of $f\left(x\right)=\frac{1-{\displaystyle \frac{1}{x}}}{\sqrt{x}}is\frac{-x+3}{2x^2\sqrt{x}}.$

Step-by-step solution

Step 1  / 2Step 1: Given information

The given expression is$f\left(x\right)=\frac{1-{\displaystyle \frac{1}{x}}}{\sqrt{x}}.$

Simplification 

$$\begin{gathered} f\text{'}\left(x\right) \\ =\frac{d}{dx}\left(\frac{1-{\displaystyle \frac{1}{x}}}{\sqrt{x}}\right) \\ =\frac{d}{dx}\left(\frac{{\displaystyle \frac{x-1}{x}}}{\sqrt{x}}\right) \\ =\frac{d}{dx}\left(\frac{x-1}{x\sqrt{x}}\right) \\ =\frac{d}{dx}\left(\frac{x}{x\sqrt{x}}-\frac{1}{x\sqrt{x}}\right) \\ =\frac{d}{dx}\left(x^{-\frac{1}{2}}\right)-\frac{d}{dx}\left(x^{-\frac{3}{2}}\right) \\ =-\frac{1}{2x^{{\displaystyle \frac{3}{2}}}}+\frac{3}{2x^{-{\displaystyle \frac{5}{2}}}} \\ =\frac{-x+3}{2x^2\sqrt{x}} \\ Hence,f\text{'}\left(x\right)=\frac{-x+3}{2x^2\sqrt{x}}. \end{gathered}$$