If $$y = \frac{\sqrt{x+1}(x^2 - \sqrt{x}}{x\sqrt{x} + x + \sqrt{x}}) + \frac{1}{15}(3\cos 2x - 5\cos 3x)$$, then $$96y'\left(\frac{\pi}{6}\right)$$ is equal to:

$$u = \frac{(\sqrt{x}+1)\sqrt{x}(x^{3/2}-1)}{\sqrt{x}(x+\sqrt{x}+1)} = \frac{(\sqrt{x}+1)(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1} = x - 1$$

Let the second term be $$v = \frac{1}{15}(3\cos^5 x - 5\cos^3 x)$$:

$$v' = \frac{1}{15}(15\cos^4 x(-\sin x) - 15\cos^2 x(-\sin x)) = \sin x\cos^2 x(1 - \cos^2 x) = \sin^3 x\cos^2 x$$

$$y = u + v$$: $$y' = 1 + \sin^3 x\cos^2 x$$

$$y'\left(\frac{\pi}{6}\right) = 1 + \sin^3\left(\frac{\pi}{6}\right)\cos^2\left(\frac{\pi}{6}\right) = 1 + \left(\frac{1}{2}\right)^3 \left(\frac{\sqrt{3}}{2}\right)^2 = 1 + \frac{1}{8}\cdot\frac{3}{4} = 1 + \frac{3}{32} = \frac{35}{32}$$

$$96 y'\left(\frac{\pi}{6}\right) = 96 \cdot \frac{35}{32} = 3 \cdot 35 = 105$$