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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