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

2 Answers
Mar 16, 2018

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

Explanation:

#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)#

Mar 16, 2018

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

Explanation:

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

So

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