18. [-/3 Points] DETAILS MY NOTES SCALCET9 3.2.045. Suppose that \( f(5)=3, f^{\prime}(5)=4, g(5)=-6 \), and \( g^{\prime}(5)=7 \). Find the following values. (a) \( (f g)^{\prime}(5) \) (b) \( \left(\frac{f}{g}\right)^{\prime}(5) \) (c) \( \left(\frac{g}{f}\right)^{\prime}(5) \)

To find the values for each part, we can use the product rule and quotient rule from calculus. For part (a): Using the product rule \((fg)' = f'g + fg'\), we have: \[ (fg)'(5) = f'(5)g(5) + f(5)g'(5) \] Substituting the known values: \[ = 4 \cdot (-6) + 3 \cdot 7 = -24 + 21 = -3 \] For part (b): Using the quotient rule \(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\), we have: \[ \left(\frac{f}{g}\right)'(5) = \frac{f'(5)g(5) - f(5)g'(5)}{g(5)^2} \] Substituting the values: \[ = \frac{4 \cdot (-6) - 3 \cdot 7}{(-6)^2} = \frac{-24 - 21}{36} = \frac{-45}{36} = -\frac{5}{4} \] For part (c): Using the quotient rule again \(\left(\frac{g}{f}\right)' = \frac{g'f - gf'}{f^2}\), we find: \[ \left(\frac{g}{f}\right)'(5) = \frac{g'(5)f(5) - g(5)f'(5)}{f(5)^2} \] Substituting the values: \[ = \frac{7 \cdot 3 - (-6) \cdot 4}{3^2} = \frac{21 + 24}{9} = \frac{45}{9} = 5 \] Final results: (a) \((fg)'(5) = -3\) (b) \(\left(\frac{f}{g}\right)'(5) = -\frac{5}{4}\) (c) \(\left(\frac{g}{f}\right)'(5) = 5\)