Suppose \( f \) and \( g \) are both defined and differentiable at \( x=0 \). Suppose that \( f(0)=-5, g(0)=1, f^{\prime}(0)=7 \) and \( g^{\prime}(0)=3 \). Then \( \left(\frac{g}{f}\right)^{\prime}(0)=\frac{A}{B} \). The value of \( A \) is

To find \( \left(\frac{g}{f}\right)^{\prime}(0) \), we can use the quotient rule for differentiation. According to the quotient rule, if \( u = g(x) \) and \( v = f(x) \), then \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. \] Substituting \( u = g \) and \( v = f \), we find: \[ \left(\frac{g}{f}\right)'(0) = \frac{g'(0)f(0) - g(0)f'(0)}{(f(0))^2}. \] From the problem, we have: - \( g(0) = 1 \) - \( f(0) = -5 \) - \( g'(0) = 3 \) - \( f'(0) = 7 \) Now we can substitute these values into the quotient rule formula: \[ \left(\frac{g}{f}\right)'(0) = \frac{3 \cdot (-5) - 1 \cdot 7}{(-5)^2}. \] Calculating the numerator: \[ 3 \cdot (-5) = -15, \quad \text{and} \quad -15 - 7 = -22. \] Then for the denominator: \[ (-5)^2 = 25. \] Now putting it all together: \[ \left(\frac{g}{f}\right)'(0) = \frac{-22}{25}. \] This gives us \( A = -22 \) and \( B = 25 \). Therefore, the value of \( A \) is \[ \boxed{-22}. \]