Where does f(x) concave up

f(x)=(x+12)*ln(x+1)
Where does this concave up? x>-1

I think f '(x)=(x+12)/(x+1)+ln(x+1) But then I guess I mess up getting f ''(x)
Cause I get (x+13)/(x+1)^2+(1/x+1) and set it equal to zero and get 7 which isn't right...
Please show how you get it.

Thanks!

f"(x) =[(x+1)-(x+12)]/(x+1)^2 + 1/(x+1) = (x-10)/(x+1)^2 = 0 for x = 10

Watch the signs! d/dx (f/g) = (f'g - fg')/g^2, not (f'g+fg')/g^2

Calculate f "(x) and find x such that f "(x) >= 0 using a sign table. That is the interval where f(x) is concave up, which means the graph lies above the tangent line at each of those points.

f '(x)=(x+12)/(x+1)+ln(x+1)
f '' (x) = (x+1 - x - 12 )/(x+1)^2 +1 / (x+1) = ( -11 +x + 1)/(x+1)^2 then x = 10