How do you find the derivative of ${sin}^{2} (2 x) + sin (2 x + 1)$ $sin ^2 (2x) + sin (2x+1)$ ?

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How do you find the derivative of ${sin}^{2} (2 x) + sin (2 x + 1)$ $sin ^2 (2x) + sin (2x+1)$ ?

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Answer 1 · 2016-03-09T00:43:20

It is

$\frac{d}{d x} [{sin}^{2} (2 x) + sin (2 x + 1)] = 2 sin 2 x (\frac{d}{d x} (sin 2 x)) + \frac{d}{d x} [sin (2 x + 1)] = 4 \cdot sin 2 x \cdot cos 2 x + 2 \cdot cos (2 x + 1)$ $d/dx [sin ^2 (2x) + sin (2x+1)]=2sin2x(d/dx(sin2x))+d/dx[sin(2x+1)] =4*sin2x*cos2x+2*cos(2x+1)$

Finally

$\frac{d}{d x} [{sin}^{2} (2 x) + sin (2 x + 1)] = 4 \cdot sin 2 x \cdot cos 2 x + 2 \cdot cos (2 x + 1)$ $d/dx [sin ^2 (2x) + sin (2x+1)]=4*sin2x*cos2x+2*cos(2x+1)$

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How do you find the derivative of ${sin}^{2} (2 x) + sin (2 x + 1)$ $sin ^2 (2x) + sin (2x+1)$ ?

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