## Calculus: Early Transcendentals 8th Edition

a. $h'(4) = -6$ b. $h'(4) = 24$ c. $h'(4) = \frac{36}{25}$ d. $h'(4) = -\frac{36}{49}$
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. $h(x) = \frac{g(x)}{f(x)+g(x)}$ $h'(x) = \frac{(f(x)+g(x))(g'(x)) - g(x)((f'(x)+g'(x))}{(f(x)+g(x))^{2}}$ $h'(4) = \frac{(f(4)+g(4))(g'(4)) - g(4)((f'(4)+g'(4))}{(f(4)+g(4))^{2}}$ $h'(4) = \frac{(2 +5)(-3) - 5(6+(-3))}{(2+5)^{2}}$ $h'(4) = \frac{(7)(-3) - 5(6-3))}{(7)^{2}}$ $h'(4) = \frac{(7)(-3) - 5(3)}{(7)^{2}}$ $h'(4) = \frac{-21 - 15}{49}$ $h'(4) = -\frac{36}{49}$