Question

d/dx (cosx+sinx)^2

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Answer 1

Step-by-step explanation:

$\frac{d}{dx} ( \cos \: x + \sin \: x) ^{2} \\ \\ = 2( \cos \: x + \sin \: x) \frac{d}{dx} ( \cos \: x + \sin \: x) \\ \\ = 2( \cos \: x + \sin \: x) (\frac{d}{dx} \cos \: x + \frac{d}{dx}\sin \: x) \\ \\ = 2( \cos \: x + \sin \: x) ( - \sin \: x + \cos \: x) \\ \\ = 2( \cos \: x + \sin \: x) (\cos \: x - \sin \: x ) \\ \\ = 2(\cos^{2} \: x - \sin^{2} \: x ) \\ \\ = 2 \cos \: 2x \\ \\ thus \\ \\ \frac{d}{dx} ( \cos \: x + \sin \: x) ^{2} = 2\cos2x$