Jambmaths question:
$\begin{align} & \text{If }y={{(2x+2)}^{3}},\text{ find }\frac{dy}{dx} \\ & (A)\text{ }6{{(2x+2)}^{2}}\text{ }(B)\text{ }3{{(2x+2)}^{2}}\text{ }(C)\text{ }6(2x+2)\text{ }(D)\text{ }3(2x+2) \\\end{align}$
Jamb Maths Solution:
$\begin{align} & y={{(2x+2)}^{3}} \\ & \text{Let }u=2x+2 \\ & y={{u}^{3}} \\ & \frac{dy}{du}=3{{u}^{2}} \\ & \frac{du}{dx}=2 \\ & \text{Using chain rule }\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx} \\ & \frac{dy}{dx}=3{{u}^{2}}\times 2=6{{u}^{2}}=6{{(2x+2)}^{2}} \\\end{align}$
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