Answer
a. $h'(4) = -6$
b. $h'(4) = 24$
c. $h'(4) = \frac{36}{25}$
d. $h'(4) = -\frac{36}{49}$
Work Step by Step
From the problem we know that $f'(4) = 2, g(4) = 5, f'(4) = 6,$ and $g'(4) = -3$. So now we have to find $h'(4)$ in all the problems.
a. $h(x) = 3f(x) + 8g(x)$
$h'(x) = 3f'(x) + 8g'(x)$
$h'(4) = 3f'(4) + 8g'(4)$
$h'(4) = 3(6) + 8(-3)$
$h'(4) = 18 - 24$
$h'(4) = -6$
b. $h(x) = f(x)g(x)$
$h'(x) = f'(x)g(x)+ g'(x)f(x)$
$h'(4) = f'(4)g(4)+ g'(4)f(4)$
$h'(4) = 6(5) + -3(2)$
$h'(4) = 30 - 6$
$h'(4) = 24$
c. $h(x) = \frac{f(x)}{g(x)}$
$h'(x) = \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^{2}}$
$h'(4) = \frac{f'(4)g(4) - g'(4)f(4)}{(g(4))^{2}}$
$h'(4) = \frac{6(5) - (-3)(2)}{(5)^{2}}$
$h'(4) = \frac{30 -(-6)}{25}$
$h'(4) = \frac{30 + 6}{25}$
$h'(4) = \frac{36}{25}$
d. $h(x) = \frac{g(x)}{f(x)+g(x)}$
$h'(x) = \frac{(f(x)+g(x))(g'(x)) - g(x)((f'(x)+g'(x))}{(f(x)+g(x))^{2}}$
$h'(4) = \frac{(f(4)+g(4))(g'(4)) - g(4)((f'(4)+g'(4))}{(f(4)+g(4))^{2}}$
$h'(4) = \frac{(2 +5)(-3) - 5(6+(-3))}{(2+5)^{2}}$
$h'(4) = \frac{(7)(-3) - 5(6-3))}{(7)^{2}}$
$h'(4) = \frac{(7)(-3) - 5(3)}{(7)^{2}}$
$h'(4) = \frac{-21 - 15}{49}$
$h'(4) = -\frac{36}{49}$
