Answer:
(fg)'(5) = -40
(f/g)'(5) = -68/36
(g/f)'(5)= 17
Step-by-step explanation:
(fg)'(5) =
Applying the product derivative:
[tex]fg(5) = f^{\prime}(5)g(5) + g^{\prime}(5)f(5) = 9*(-6) + 7*2 = -54 + 14 = -40[/tex]
(f/g)'(5) =
Applying the quotient derivative
[tex](\frac{f}{g})^{\prime}(5) = \frac{f^{\prime}(5)g(5) - g^{\prime}(5)f(5)}{g(5)^2} = \frac{9*(-6) - 7*2}{6^2} = -\frac{68}{36}[/tex]
(g/f)'(5)=
[tex](\frac{g}{f})^{\prime}(5) = \frac{g^{\prime}(5)f(5) - f^{\prime}(5)g(5)}{f^{\prime}(5)^2} = \frac{7*2 - 9*(-6)}{2^2} = \frac{68}{4} = 17[/tex]