To derive this function, we'll only need the power rule.
Power rule is expressed with the following function:
[tex]\frac{d}{dx} x^n = n \cdot x^{n-1}[/tex]
We can derive each of the terms individually. Rewrite the problem:
[tex]\frac{d}{dx} [\frac{1}{\sqrt{x}} - 3.2x^{-2} + x] = (\frac{d}{dx} \frac{1}{\sqrt{x}}) - (\frac{d}{dx} 3.2x^{-2}) + (\frac{d}{dx} x) [/tex]
For the first term, we'll rewrite the term, then use the power rule (note that square roots can be rewritten as any number to the 1/2 power):
[tex]\frac{d}{dx} \frac{1}{\sqrt{x}} = \frac{d}{dx} x^{-\frac{1}{2}} = -\frac{1}{2} x^{-\frac{3}{2}} = -\frac{1}{2x^{\frac{3}{2}}} [/tex]
For the second term, we'll use power rule:
[tex]\frac{d}{dx} 3.2x^{-2} = -6.4x^{-3} = -\frac{6.4}{x^3} = -\frac{32}{5x^3}[/tex]
The third term is simple:
[tex]\frac{d}{dx} x = 1[/tex]
The equation should now look like this, and result in your answer:
[tex]\frac{d}{dx} [\frac{1}{\sqrt{x}} - 3.2x^{-2} + x] = -\frac{1}{2x^{\frac{3}{2}}} - (-\frac{32}{5x^3}) + 1 = \boxed{-\frac{1}{2x^{\frac{3}{2}}} + \frac{32}{5x^3} + 1}[/tex]
