Differentiate #y=sqrt(x^2)+sqrt(4x)-sqrt3 ?

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Answer 1 · 2018-03-16T21:10:37

#y' = x/sqrt(x^2) + 1/sqrt(x)#

#y=\sqrt(x^2)+\sqrt(4x)-\sqrt(3)#

#y' = 1/2(x^2)^(-1/2)(2x) + 1/2(4x)^(-1/2)(4)-0#

#y' = x/sqrt(x^2) + 2/\sqrt(4x)#

#y' = x/sqrt(x^2) + 1/sqrt(x)#

Answer 2 · 2018-03-16T21:14:32

#dy/dx=(absx)/x+1/sqrtx#

#y=sqrt(x^2)+sqrt (4x)-sqrt3=absx+sqrt4 x^(1/2)-sqrt3#

So

#dy/dx=absx/x+sqrt4/(2sqrtx)=(absx)/x+1/sqrtx#