Jambmaths question:
If $y={{x}^{2}}-\tfrac{1}{x},$find $\tfrac{dy}{dx}$
Option A:
$2x-\tfrac{1}{{{x}^{2}}}$
Option B:
$2x+{{x}^{2}}$
Option C:
$2x-{{x}^{2}}$
Option D:
$2x+\tfrac{1}{{{x}^{2}}}$
Jamb Maths Solution:
$\begin{align} & y={{x}^{2}}-\frac{1}{x}={{x}^{2}}-{{x}^{-1}} \\ & \frac{dy}{dx}=2x-(-1)({{x}^{-1-1}}) \\ & \frac{dy}{dx}=2x+{{x}^{-2}} \\ & \frac{dy}{dx}=2x+\frac{1}{{{x}^{2}}} \\\end{align}$
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