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y = (x^2)(sin x)(2x)(cos x) - 2sin x

y' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'

y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'

y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)

y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)

y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos x

y' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)

y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos x

y' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos x

y' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x

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Q: Find the derivative of y x2 sin x 2xcos x - 2sin x?

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(/) = theta sin 2(/) = 2sin(/)cos(/)

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The derivative of sin(x) is cos(x).

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It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!

because sin(2x) = 2sin(x)cos(x)

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