How do you differentiate #cossqrtxsqrt(cosx)#?

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Answer 1 · 2016-12-30T14:53:51

#-(1/2)((sqrtx sin x cos sqrtx+cos xsin sqrt x)/(sqrt(x cos x)))#,
#x in ( 2kpi, (2k+1/2)pi), k = 0, 1, 2, 3, ...#

To make the function real and differentiable,

#x>0#, and further, #cos x >0#, giving #x > 0# in open #Q_1#.

In brief, #x in ( 2kpi, (2k+1/2)pi). k = 0, 1, 2, 3, ...#.

Then, for this piecewise differentiation,

#((cos sqrt x)(sqrt cos x))'#

#=(cos sqrt x)( sqrtcos x)'+(cos sqrtx)(sqrt cos x)'#

#=((cos sqrtx)(1/(2(sqrt cos x))(-sin x))+((-sin sqrtx(1/(2sqrtx))(sqrtcos x)#

#=-(1/2)((sqrtx sin x cos sqrtx+cos xsin sqrt x)/(sqrt(x cos x)))#