Find h'(2), given that f(2)= -3, g(2)= 4, f'(2)= -2, and g'(2)= 7
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)'
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
Is there supposed to be a 1 + in front of the f'(2)?
((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.
To find h'(2), we need to use the quotient rule for differentiation since h(x) is defined as the ratio of two functions, g(x) and 1 + f(x). The quotient rule states that if h(x) = g(x)/f(x), then h'(x) = (g'(x)f(x) - g(x)f'(x))/[f(x)]^2.
In this case, the formula for h'(x) is different because the order of g(x) and f(x) is reversed in the denominator. So, using the quotient rule, the formula to find h'(x) is:
h'(x) = (f(x)g'(x) - g(x)f'(x))/[g(x)]^2
Now, substitute the given values into the formula to find h'(2):
h'(2) = (f(2)g'(2) - g(2)f'(2))/[g(2)]^2
Substituting the given values:
h'(2) = (-3 * 7 - 4 * -2)/(4)^2
Simplifying the numerator:
h'(2) = (-21 + 8)/(16)
Combining like terms:
h'(2) = -13/16
Therefore, h'(2) = -13/16.