Question

EVALUATE..............

Answer 1

Answer:

0

Step-by-step explanation:

$FORMULAE\\\\cos^2x+sin^2x=1\\\\sin2x=2sinxcosx\\\\(a+b)^2=a^2+b^2+2ab\\\\\frac{d}{dx}(k)=0\\\\given\\\\\frac{d}{dx}(\frac{sinx+cosx}{\sqrt{1+sin2x}})\\\\\\=\frac{d}{dx}(\frac{sinx+cosx}{\sqrt{(cosx+sinx)^2}})\\\\\\=\frac{d}{dx}(\frac{sinx+cosx}{|cosx+sinx|})\\\\\\=\frac{d}{dx}(\frac{cosx+sinx}{|cosx+sinx|})\\\\\\=\frac{d}{dx}|1|\\\\\\=0$

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Answer 2

MARK AS BRAINLIEST...............♥♥