Question 37

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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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