Question 18

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Jambmaths question: 

If $y={{x}^{2}}+\sqrt{x}$ , find $\frac{dy}{dx}$

Option A: 

$2x-\tfrac{1}{2}{{x}^{-\tfrac{1}{2}}}$

Option B: 

$2x-{{x}^{-\tfrac{1}{2}}}$

Option C: 

$2x+\tfrac{1}{2}{{x}^{-\tfrac{1}{2}}}$

Option D: 

$2x+{{x}^{\tfrac{1}{2}}}$

Jamb Maths Solution: 

$\begin{align}  & y={{x}^{2}}+\sqrt{x}={{x}^{2}}+{{x}^{\tfrac{1}{2}}} \\ & \frac{dy}{dx}=2x+\tfrac{1}{2}{{x}^{-\tfrac{1}{2}}} \\\end{align}$

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