Question 40

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Jambmaths question: 

If $y={{(2x+1)}^{3}},\text{ find }\frac{dy}{dx}$

Jamb Maths Solution: 

$\begin{align}  & \text{Using the generalized power rule which state that given} \\ & y={{\left[ f(x) \right]}^{n}}\text{ then }\frac{dy}{dx}=n\cdot {{\left[ f(x) \right]}^{n-1}}\cdot {{f}^{1}}(x) \\ & y={{(2x+1)}^{3}} \\ & \frac{dy}{dx}=3\cdot {{(2x+1)}^{3-1}}\times \tfrac{d}{dx}(2x+1) \\ & \frac{dy}{dx}=3\cdot {{(2x+1)}^{2}}(2)=6{{(2x+1)}^{2}} \\\end{align}$

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