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Tuition/Class XI-XII Tuition (PUC)

Derivative of sin3x+cos4x?

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3cos3x - 4sin4x

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3cos3x - 4sin4x

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D(sin3X) = 3cos3X. D(cos4X) = -4sin4X. D(sin3x+cos4x) = 3 cos 3x - 4 sin 4x.

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3Cos3x - 4Sin4x purpose of asking this question is to know who knows better.

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dx[sin3x+cos4x] = 3cos3x-4sin4x.

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3cos3x - 4sin4x.

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3cos3x - 4sin4x.

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d/dx (sinx)= cosx and d/dx (cosx) = -sinx, we can compute this problem from these properties. first find d/dx (sin3x) = cos3x * d/dx (3x) = cos3x * 3 and d/dx (cos4x) = -sin4x d/dx(4x) = - sin4x * 4 so Ans is 3cos3x- 4 sin4x

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3cos3x - 4sin4x.

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3cosx - 4sinx

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