Question 15

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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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