Suppose that \( f(4)=4, f^{\prime}(4)=-4, g(4)=4 \), and \( g^{\prime}(4)=2 \). If \( F(x)=(f \circ g)(x) \) and \( G(x)=(g \circ f)(x) \), find \( F^{\prime}(4)+G^{\prime}(4) \). Answer
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Bonus Knowledge
To solve for \( F'(4) + G'(4) \) where \( F(x) = (f \circ g)(x) \) and \( G(x) = (g \circ f)(x) \), we use the chain rule to find the derivatives. First, we compute \( F'(x) = (f \circ g)'(x) \): \[ F'(x) = f'(g(x)) \cdot g'(x) \] Now, evaluate \( F'(4) \): \[ F'(4) = f'(g(4)) \cdot g'(4) \] Given that \( g(4) = 4 \), we get: \[ F'(4) = f'(4) \cdot g'(4) \] With \( f'(4) = -4 \) and \( g'(4) = 2 \): \[ F'(4) = (-4) \cdot 2 = -8 \] Next, we compute \( G'(x) = (g \circ f)'(x) \): \[ G'(x) = g'(f(x)) \cdot f'(x) \] Now, evaluate \( G'(4) \): \[ G'(4) = g'(f(4)) \cdot f'(4) \] Given that \( f(4) = 4 \): \[ G'(4) = g'(4) \cdot f'(4) \] With \( g'(4) = 2 \) and \( f'(4) = -4 \): \[ G'(4) = 2 \cdot (-4) = -8 \] Finally, we find \( F'(4) + G'(4) \): \[ F'(4) + G'(4) = -8 + (-8) = -16 \] Thus, the answer is: \[ \boxed{-16} \]