Help with 2nd derivative

x/(x+1)^2
solve for the second derivative, at x = 5

is it 1/216...?
please help

Differentiate the function twice, using quotient and chain rules...

Quotient - d/dx f/g = (gf' - fg')/g²
Chain - d/dx f(g) = f'(g)g'

dy/dx = ((x + 1)²(1) - x(2(x + 1)))/((x + 1)²)²
= (x + 1)(x + 1 - 2x)/(x + 1)^4
= (-x + 1)/(x + 1)³

d²y/dx² = ((x + 1)³(-1) - (-x + 1)(3(x + 1)²))/((x + 1)³)²
= (x + 1)²(-x - 1 + 3(x - 1))/(x + 1)^6
= (-x - 1 + 3x - 3)/(x + 1)^4
= (2x - 4)/(x + 1)^4

Finally, if x = 5, then we obtain:

d²y/dx² = (2(5) - 4)/(5 + 1)^4
= 6/(6)^4
= 1/6³
= 1/216 [You are correct!]

I hope this helps!

x/(x+1)^2
x(x+1)^(-2)
(fg)' = f'g + g'f
(1)(x+1)^(-2)dx + x(-2)(x+1)^(-3)dx
(x+1)^(-2)dx - 2x(x+1)^(-3)dx
(-2)(x+1)^(-3)dx^2- 2(1)(x+1)^(-3)dx^2 - 2x (-3)(x+1)^(-4)dx^2
-2(x+1)^(-3)dx^2 - 2(x+1)^(-3)dx^2 + 6x(x+1)^(-4)dx^2
-4(x+1)^(-3)dx^2 +6x(x+1)^(-4)dx^2
-4/(x+1)^3dx^2 + 6x/(x+1)^4dx^2
then we substitute x with 5
-4/(5+1)^3 + 6(5)/(5+1)^4
-4/6^3 + 6(5)/6^4
-4/6^3 + 5/6^3
(-4 +5)/216
=1/216
yup :)

f(x) = 1/(x+1) - 1/(x+1)^2
f '(x) = -1/(x+1)^2 + 2/(x+1)^3
f "(x) = 2/(x+1)^3 - 6/(x+1)^4
f "(5) = ... = 1/216