Suppose that h(x) = f(x) cdot g(x) and F(x) = | vedclass

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(B) Given $h(x) = f(x)g(x)$. Using the product rule,$h'(x) = f'(x)g(x) + g'(x)f(x)$.At $x = 2$,$h'(2) = f'(2)g(2) + g'(2)f(2) = (-2)(5) + (4)(3) = -10

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$F'(2) = 22 h'(2)$

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(B) Given $h(x) = f(x)g(x)$. Using the product rule,$h'(x) = f'(x)g(x) + g'(x)f(x)$.At $x = 2$,$h'(2) = f'(2)g(2) + g'(2)f(2) = (-2)(5) + (4)(3) = -10 + 12 = 2$.Given $F(x) = f(g(x))$. Using the chain rule,$F'(x) = f'(g(x)) \cdot g'(x)$.At $x = 2$,$F'(2) = f'(g(2)) \cdot g'(2) = f'(5) \cdot 4 = 11 \cdot 4 = 44$.Now,comparing $F'(2)$ and $h'(2)$,we have $F'(2) = 44$ and $h'(2) = 2$.Therefore,$F'(2) = 22 \cdot h'(2)$.

Suppose that $h(x) = f(x) \cdot g(x)$ and $F(x) = f(g(x))$,where $f(2) = 3$,$g(2) = 5$,$g'(2) = 4$,$f'(2) = -2$,and $f'(5) = 11$. Then:

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