Jambmaths question:
Find the derivative o the function $y=2{{x}^{2}}(2x-1)$ at the point x = –1
Option A:
18
Option B:
16
Option C:
–4
Option D:
–6
Jamb Maths Solution:
$\begin{align} & y=2{{x}^{2}}(2x-1) \\ & \frac{dy}{dx}=U\frac{dv}{dx}+V\frac{du}{dx} \\ & \frac{dy}{dx}=(2{{x}^{2}})\frac{d}{dx}(2x-1)+(2x-1)\frac{d}{dx}(2{{x}^{2}}) \\ & \frac{dy}{dx}=(2{{x}^{2}})(2)+(2x-1)(4x)=4{{x}^{2}}+8{{x}^{2}}-4x \\ & \frac{dy}{dx}=12{{x}^{2}}-4x \\ & \frac{dy}{dx}\left| _{x=-1} \right.=12{{(-1)}^{2}}-4(-1)=12+4=16 \\\end{align}$
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