How do you find the second derivative of #z=xsqrt(x+1)#?

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Answer 1 · 2017-09-11T16:35:06

#(d^2z)/dx^2 = (3x+4) /(4(x+1)^(3/2))#

Calculate the first derivative using the product rule:

#(dz)/dx = d/dx ( xsqrt(x+1) ) = (d/dx x) sqrt(x+1) + x (d/dx sqrt(x+1))#

#(dz)/dx = sqrt(x+1) + x/(2sqrt(x+1))#

Simplifying:

#(dz)/dx = (2(x+1) + x)/(2sqrt(x+1)) = (3x+2)/(2sqrt(x+1))#

Differentiate again using the quotient rule:

#(d^2z)/dx^2 = d/dx ((3x+2)/(2sqrt(x+1)))#

#(d^2z)/dx^2 = ((2sqrt(x+1))d/dx (3x+2) - (3x+2)(d/dx 2sqrt(x+1)) )/(2sqrt(x+1))^2#

#(d^2z)/dx^2 = ((6sqrt(x+1)) - (3x+2)/sqrt(x+1)) /(4(x+1))#

#(d^2z)/dx^2 = (6x+6 - 3x-2) /(4(x+1)^(3/2))#

#(d^2z)/dx^2 = (3x+4) /(4(x+1)^(3/2))#