Question 6

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Jambmaths question: 

Find the derivative of $y=\sin (2{{x}^{3}}+3x-4)$

Option A: 

$-\cos (2{{x}^{2}}+3x-4)$

Option B: 

$-(6{{x}^{2}}+3)\cos (2{{x}^{2}}+3x-4)$

Option C: 

$cos (2{{x}^{2}}+3x-4)$

Option D: 

$(6{{x}^{2}}+3)\cos (2{{x}^{3}}+3x+4)$

Jamb Maths Solution: 

$\begin{align}  & y=\sin (2{{x}^{3}}+3x-4) \\ & \text{Let }u=2{{x}^{3}}+3x-4,\frac{du}{dx}=6{{x}^{2}}+3 \\ & y=\sin u,\text{ }\frac{dy}{du}=\cos u \\ & \text{Using the chain Rule} \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=\cos u\times (6{{x}^{2}}+3)=(6{{x}^{2}}+3)\cos u \\ & \frac{dy}{dx}=(6{{x}^{2}}+3)\cos (2{{x}^{3}}+3x-4) \\\end{align}$

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