Find h'(2), given that f(2)= -3, g(2)= 4, f'(2)= -2, and g'(2)= 7

Question

d) h(x)= g(x)/(1 + f(x))

What would the formula be? Normally it is (f/g)'= (gf' - fg')/g^2 But in this case, it is (g/f)'

Answer 1

That would be similar to the quotient case, where the numerator is u=g(x), and the denominator v=(1+f(x))
Applying the quotient rule,
d(u/v)=(vdu-udv)/v²

so
(g(x)/(1+f(x))'
=((1+f(x))g'(x)-g(x)f'(x))/(1+f(x))²

Answer 2

Is there supposed to be a 1 + in front of the f'(2)?

Answer 3

((1+f(x))g'(x)-g(x)f'(x))/(1+f(x))² is correct, but I have jumped steps because
(1+f(x))' = (0+f'(x)) = f'(x).

Sorry for the confusion.