Jambmaths question:
Find the derivative of $y={{(\tfrac{1}{3}x+6)}^{2}}$
Option A:
$2(\tfrac{1}{3}x+6)$
Option B:
$\tfrac{2}{3}(\tfrac{1}{3}x+6)$
Option C:
$\tfrac{2}{3}{{(\tfrac{1}{3}x+6)}^{2}}$
Option D:
$\tfrac{1}{3}{{(\tfrac{1}{3}x+6)}^{2}}$
Jamb Maths Solution:
$\begin{align} & y={{[f(x)]}^{n}} \\ & \text{Using the generalised derivative formula} \\ & \frac{dy}{dx}=n\times {{[f(x)]}^{n-1}}\times f'(x) \\ & y={{(\tfrac{1}{3}x+6)}^{2}} \\ & \frac{dy}{dx}=2\times [(\tfrac{1}{3}x+6)]\times \tfrac{d}{dx}(\tfrac{1}{3}x+6) \\ & \frac{dy}{dx}=2[(\tfrac{1}{3}x+6)]\times \tfrac{1}{3}=\tfrac{2}{3}(\tfrac{1}{3}x+6) \\ & \frac{dy}{dx}=\tfrac{2}{3}(\tfrac{1}{3}x+6) \\\end{align}$
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