how to differentiate cos2x, sin2x, tan2x, etc

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Solution

$\frac{d}{dx} (\cos 2 x) = - \sin 2 x \times \frac{d}{dx} [2 x] = - 2 \sin 2 x \frac{d}{dx} (\sin 2 x) = \cos 2 x \times \frac{d}{dx} [2 x] = 2 \cos 2 x \frac{d}{dx} (\tan 2 x) = \sec^{2} 2 x \times \frac{d}{dx} [2 x] = 2 \sec^{2} 2 x$

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