\[f(x) = \sqrt{x^{3}} + 1;\ \ \ x_{0} = 4:\]
\[f^{'}(x) = \left( x^{\frac{3}{2}} \right)^{'} + (1)^{'} =\]
\[= \frac{3}{2}x^{\frac{1}{2}} + 0 = \frac{3}{2}\sqrt{x};\]
\[f^{'}(4) = \frac{3}{2} \bullet \sqrt{4} = \frac{3}{2} \bullet 2 = 3;\]
\[f(4) = \sqrt{4^{3}} + 1 = \sqrt{64} + 1 =\]
\[= 8 + 1 = 9;\]
\[y = 9 + 3(x - 4) =\]
\[= 9 + 3x - 12 = 3x - 3.\]
\[Ответ:\ \ y = 3x - 3.\]