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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