A Warm up problem (1)

$\begin{aligned} \phi(x) & = \frac 1{x^2} \int_4^x \left(4t^2 - 2\phi'(x)\right) dt \\ x^2 \phi(x) & = \int_4^x \left(4t^2 - 2\phi'(x)\right) dt \\ & = \left[ \frac {4t^3}3 - 2\phi (x) \right]_4^x \\ & = \frac {4x^3-4^4}3 - 2\phi(x) + 2\phi(4) & ...(1) \end{aligned}$

Putting $x=4$ in equation 1:

$\begin{aligned} 16 \phi(4) & = \frac {4^4-4^4}3 - 2\phi(4) + 2\phi(4) \\ \implies \phi (4) & = 0 \end{aligned}$

Differentiating both sides of equation 1 with respect to $x$ :

$\begin{aligned} 2x \phi (x) + x^2 \phi'(x) & = 4x^2 - 2\phi'(x) & \small \color{#3D99F6} \text{Putting }x=4 \\ 8{\color{#3D99F6}\phi (4)} + 16\phi'(4) & = 64 - 2\phi'(4) & \small \color{#3D99F6} \text{Note that }\phi(4) = 0 \\ {\color{#3D99F6}0} + 18\phi'(4) & = 64 \\ \phi'(4) & = \frac {64}{18} \\ & \approx \boxed{3.556} \end{aligned}$