Question
If $\text{f}'\text{(x)}=\text{x}-\frac{1}{\text{x}^2}$ and $\text{f}'(1)=\frac{1}{2}$, find f'(x).

Answer

$\text{f}'\text{(x)}=\text{x}-\frac{1}{\text{x}^2}$
$\text{f}'\text{(x)}=\text{x}-\text{x}^{-2}$
$\int\text{f}'\text{(x)}\text{dx}=\int(\text{x}-\text{x}^{-2})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}-\frac{\text{x}^{-2+1}}{-2+1}+\text{C}$
$=\frac{\text{x}^2}{2}+\frac{1}{\text{x}}+\text{C}$
$\text{f}'\text{(1)}=\frac{1}{2}$ (Given)
$\Rightarrow\frac{{1}^2}{2}+\frac{1}{1}+\text{C}=\frac{1}{2}$
$\Rightarrow\text{C}=-1$
$\therefore\ \text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{1}{\text{x}}-1$

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