Find gradient of curve y=(x^3-1)ln(x^2 -1 ),at the point where x=2,state answer in 3 significant figure

ok heres my step
du/dx = (x^3-1)(3x) dv/dx= 2x
dy/dx = (x^3-1)(2x) +ln(2^2 -1 )*(x^2 -1 )(3x)
then sub x= 2
i got the wrong answer as the answer given is 22.5
thanks in advance

dy/dx = d/dx(x^3 - 1) * ln(x^2 - 1) + (x^3 - 1) * d/dx(ln(x^2 - 1))
dy/dx = 3x^2 ln(x^2 - 1) + (x^3 - 1) 2x/(x^2 - 1)
dy/dx = 3x^2 ln(x^2 - 1) + (2x^4 - 2x)/(x^2 - 1)
dy/dx(at x = 2) = 13.183 + 9.333 = 22.5