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

1 Answer
Sep 11, 2017

(d^2z)/dx^2 = (3x+4) /(4(x+1)^(3/2))d2zdx2=3x+44(x+1)32

Explanation:

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))dzdx=ddx(xx+1)=(ddxx)x+1+x(ddxx+1)

(dz)/dx = sqrt(x+1) + x/(2sqrt(x+1))dzdx=x+1+x2x+1

Simplifying:

(dz)/dx = (2(x+1) + x)/(2sqrt(x+1)) = (3x+2)/(2sqrt(x+1))dzdx=2(x+1)+x2x+1=3x+22x+1

Differentiate again using the quotient rule:

(d^2z)/dx^2 = d/dx ((3x+2)/(2sqrt(x+1)))d2zdx2=ddx(3x+22x+1)

(d^2z)/dx^2 = ((2sqrt(x+1))d/dx (3x+2) - (3x+2)(d/dx 2sqrt(x+1)) )/(2sqrt(x+1))^2d2zdx2=(2x+1)ddx(3x+2)(3x+2)(ddx2x+1)(2x+1)2

(d^2z)/dx^2 = ((6sqrt(x+1)) - (3x+2)/sqrt(x+1)) /(4(x+1))d2zdx2=(6x+1)3x+2x+14(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))