Answer
a) $-7$
b) $-\frac{5}{4}$
c) $\frac{3}{2}$
d) $0$
Work Step by Step
We find:
a) $\frac{d}{dx} [f(x)g(x)]|_{x=1} = f'(x)g(x) + f(x)g'(x)|_{x=1} = f'(1)g(1) + f(1)g'(1) = -1 -6 = -7$
b) $\frac{d}{dx} [\frac{f(x)}{g(x)}]|_{x=1} = \frac{f'(x)g(x) + f(x)g'(x)}{(g(x))^2}|_{x=1} = \frac{f'(1)g(1) + f(1)g'(1)}{(g(1))^2} = \frac{-6-(-1)}{2^2} = \frac{-5}{4}$
c) $\frac{d}{dx}[\sqrt{f(x)}]|_{x=1} = \frac{1}{2} (f(x))^{-\frac{1}{2}} f'(x)|_{x=1} = \frac{1}{2} (f(1))^{-\frac{1}{2}} f'(1) = \frac{1}{2} * 1 * 3 = \frac{3}{2}$
d) $\frac{d}{dx} [f(1)g'(1)] = \frac{d}{dx} 1 * (-1) = 0$