Question

Find
h ′(2), given that f(2) = −3, g(2) = 4, f ′(2) = −2,and g ′(2) = 5.

I only need help with part d (picture attached)

Answer 1

Answer: $-\frac{1}{2}$

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Work Shown:

Apply the quotient rule

$h(x) = \frac{g(x)}{1+f(x)}\\\\h(x) = \frac{A}{B}\\\\h'(x) = \frac{A'*B-A*B'}{B^2}\\\\h'(x) = \frac{g'(x)*(1+f(x))-g(x)*(1+f(x))'}{(1+f(x))^2}\\\\h'(x) = \frac{g'(x)*(1+f(x))-g(x)*f'(x)}{(1+f(x))^2}\\\\h'(2) = \frac{g'(2)*(1+f(2))-g(2)*f'(2)}{(1+f(2))^2}\\\\h'(2) = \frac{5*(1-3)-4*(-2)}{(1-3)^2}\\\\h'(2) = \frac{-2}{4}\\\\h'(2) = -\frac{1}{2}$