Let \( F(x)=f\left(x^{5}\right) \) and \( G(x)=(f(x))^{5} \) You also know that \( a^{4}=7, f(a)=3, f^{\prime}(a)=6, f^{\prime}\left(a^{5}\right)=8 \) Then \( F^{\prime}(a)=\square \) and \( G^{\prime}(a)= \)

To find \( F^{\prime}(a) \), we use the chain rule on \( F(x) = f(x^5) \): \[ F^{\prime}(x) = f^{\prime}(x^5) \cdot \frac{d}{dx}(x^5) = f^{\prime}(x^5) \cdot 5x^4. \] Now, substituting \( x = a \): \[ F^{\prime}(a) = f^{\prime}(a^5) \cdot 5a^4. \] Given \( a^4 = 7 \), \( f^{\prime}(a^5) = 8 \): \[ F^{\prime}(a) = 8 \cdot 5 \cdot 7 = 280. \] For \( G^{\prime}(a) \), we apply the chain and product rules on \( G(x) = (f(x))^5 \): \[ G^{\prime}(x) = 5(f(x))^4 \cdot f^{\prime}(x). \] Substituting \( x = a \): \[ G^{\prime}(a) = 5(f(a))^4 \cdot f^{\prime}(a). \] Given \( f(a) = 3 \) and \( f^{\prime}(a) = 6 \): \[ G^{\prime}(a) = 5(3)^4 \cdot 6 = 5 \cdot 81 \cdot 6 = 2430. \] So, \( F^{\prime}(a) = 280 \) and \( G^{\prime}(a) = 2430 \).