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

Question

Charles Arocha · Accepted Answer

#(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))#

Charles Arocha · Answer

undefined To find the second derivative of $ z = x\sqrt{x+1} $, follow these steps:

1. Start by finding the first derivative of $ z $ with respect to $ x $ using the product rule.
   
   $$ \frac{dz}{dx} = \sqrt{x+1} + x\left(\frac{1}{2\sqrt{x+1}}\right) $$

2. Simplify the expression:
   
   $$ \frac{dz}{dx} = \sqrt{x+1} + \frac{x}{2\sqrt{x+1}} $$

3. Then differentiate $ \frac{dz}{dx} $ with respect to $ x $ to find the second derivative.

$$ \frac{d^2z}{dx^2} = \frac{1}{2\sqrt{x+1}} + \frac{1}{2\sqrt{x+1}} - \frac{x}{4(x+1)^{\frac{3}{2}}} $$

4. Combine like terms and simplify further:

$$ \frac{d^2z}{dx^2} = \frac{1}{\sqrt{x+1}} - \frac{x}{2(x+1)^{\frac{3}{2}}} $$

So, the second derivative of $ z $ with respect to $ x $ is $ \frac{1}{\sqrt{x+1}} - \frac{x}{2(x+1)^{\frac{3}{2}}} $.