84. Let \( r(x)=f(g(h(x))) \), where \( h(1)=2, g(2)=3, h^{\prime}(1)=4 \), \( g^{\prime}(2)=5 \), and \( f^{\prime}(3)=6 \). Find \( r^{\prime}(1) \). 85. Suppose \( g \) is a twice differentiable function (the first derivative \( g^{\prime} \) and the second derivative \( g^{\prime \prime} \) exist) and \( f(x)=x g\left(x^{2}\right) \). Find \( f^{\prime \prime} \) in terms of \( g, g^{\prime} \), and \( g^{\prime \prime} \). 86. Suppose \( F(x)=f(3 f(4 f(x))) \), where \( f(0)=0 \) and \( f^{\prime}(0)=2 \). Find \( F^{\prime}(0) \).

To find \( r'(1) \) for the function \( r(x) = f(g(h(x))) \), we can apply the chain rule. The derivative will be \( r'(1) = f'(g(h(1))) \cdot g'(h(1)) \cdot h'(1) \). Plugging in the values, we have \( h(1) = 2 \), \( g(2) = 3 \), and the derivatives \( h'(1) = 4 \) and \( g'(2) = 5 \). Thus, \( r'(1) = f'(3) \cdot 5 \cdot 4 = 20f'(3) \). Given \( f'(3) = 6 \), the final value is \( r'(1) = 20 \cdot 6 = 120 \). To find \( f''(x) \) for the function \( f(x) = xg(x^2) \), we first need to compute the first derivative using the product rule: \( f'(x) = g(x^2) + 2x^2g'(x^2) \). For the second derivative, we differentiate \( f'(x) \) again: \[ f''(x) = g'(x^2) \cdot 2x + 2x^2g''(x^2) \cdot 2x + 2g'(x^2) = 2g'(x^2)x + 4x^3g''(x^2) + 2g'(x^2) = 2g'(x^2)(1 + 2x^2) + 4x^3g''(x^2). \] Now, putting it all together gives you \( f''(x) \) expressed neatly using \( g, g', \) and \( g'' \). To find \( F'(0) \) for the function \( F(x) = f(3f(4f(x))) \) and the conditions \( f(0) = 0 \) and \( f'(0) = 2 \), we first apply the chain rule. The derivative is \( F'(0) = f'(3f(4f(0))) \cdot 3f'(4f(0)) \cdot 4f'(0) \). Since \( f(0) = 0 \), we have \( F'(0) = f'(0) \cdot 3f'(0) \cdot 4f'(0) \). Substituting \( f'(0) = 2 \) gives: \[ F'(0) = 2 \cdot 3 \cdot 2 \cdot 4 = 48. \] Thus, you can see how derivatives intertwine like a fun rollercoaster of calculations!