f(2)=-5

f'(2)=2

f'(2)=limit[f(2+t)-f(2)]/t

t->0

g(2)=6

g'(2)=5

g'(2)=limit[g(2+t)-g(2)]/t

t->0

h(x)=2f(X)-2g(x)

h'(x)=limit[h(x+t)-h(x)]/t

t->0

but

h(2+t)=2f(2+t)-2g(2+t)

and h(2)=2f(2)-2g(2)

therfor

h'(x)=limit[2f(2+t)-2g(x+t)-(2f(x)-2g(x))]/t

t->0

or

h'(2)= 2xlimit[f(2+t)-f(2)]/t-2xlimit[g(2+t)-g(2)]/t

t->0 t->0

putting the values we get

h'(2)= 2[2-5]

=-6

note : due to functional equation we don't need the value of f(2) and g(2)