Quppose that \( f(1)=1, f^{\prime}(1)=5, g(1)=1 \), and \( g^{\prime}(1)=-6 \). If \( F(x)=(f \circ g)(x) \) and \( G(x)=(g \circ f)(x) \), find \( F^{\prime}(1)+G^{\prime}(1) \).

To find \( F'(1) + G'(1) \), we will need to apply the chain rule to \( F(x) = (f \circ g)(x) \) and \( G(x) = (g \circ f)(x) \). First, let's compute \( F'(x) \). \[ F'(x) = (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \] Now, we can evaluate \( F'(1) \): \[ F'(1) = f'(g(1)) \cdot g'(1) \] Given \( g(1) = 1 \), we have: \[ F'(1) = f'(1) \cdot g'(1) \] Substituting the known values: \[ F'(1) = 5 \cdot (-6) = -30 \] Next, we compute \( G'(x) \). \[ G'(x) = (g \circ f)'(x) = g'(f(x)) \cdot f'(x) \] Now, we can evaluate \( G'(1) \): \[ G'(1) = g'(f(1)) \cdot f'(1) \] Since \( f(1) = 1 \), we get: \[ G'(1) = g'(1) \cdot f'(1) \] Substituting in the known values: \[ G'(1) = -6 \cdot 5 = -30 \] Now we can combine \( F'(1) \) and \( G'(1) \): \[ F'(1) + G'(1) = -30 + (-30) = -60 \] Thus, the value of \( F'(1) + G'(1) \) is: \[ \boxed{-60} \]