Question

find dy/dx, when y= x cube. sin(2x) by using chain rule

Answer 1

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Answer 2

Hope u like my process
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Formula to be used
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d(uv) /dx = udv/dx + vdu/dx

d(sin2x) /dx = 2.cos2x

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y = x³. sin(2x)

Differentiating both sides by x
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$\frac{dy}{dx} = {x}^{3} \times \frac{d( \sin(2x) )}{dx} + \sin(2x) \times \frac{d( {x}^{3}) }{dx} \\ \\ or. \: \frac{dy}{dx} = {x}^{3} \times 2 \cos(2x) + \sin(2x) \times 3 {x}^{2} \\ \\ = 2 {x}^{3} \cos(2x) + 3 {x}^{2} \sin(2x)$
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