Differentiable functions problem!

If f , g, and h are differentiable functions such that:
f (0) = 1, g(0) = 2, h(0) = 3, (gh)′ (0) = 4, (hf )′ (0) = 5, and (f g)′ (0) = 6
find the value of (f gh)′ (0).

(fg)' = fg' +gf' = 2f' + g' = 6

(gh)' = gh' + hg' = 2h' + 3g' = 4

(fh)' = fh' + hf' = h' + 3f' = 5

This gives 3 equations in three variables. Solving for f', g', and h', we get.

f'(0) = 2, g'(0) = 2, h'(0) = -1

Now look at (fgh)' = f*(gh)' + (gh)*f' = 2*4 + (2*3)*1 = 14