The graph of the function with rule $\frac{k}{2(x^3+1)}$ has gradient 1 when $x=1$.Find the value of k
the answer is $\frac{-8}{3}$
I did it
find derivative of
$\frac{k}{2(x^3+1)}$
$2(3x^2)$
$6x^2$
sub $x=1$
$6(1)^2=6$
$\frac{k}{2(x^3+1)}$ $x=6$
$1=\frac{k}{12}$
$k=12$ still can't get $\frac{-8}{3}$
help me out thanks.