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

$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given below:
$f'(x)$ sign chart:
- For $x < -5$,$f'(x) > 0$
- For $-5 < x < 2$,$f'(x) < 0$
- For $2 < x < 4$,$f'(x) > 0$
- For $x > 4$,$f'(x) < 0$
From the possible graph of $y = f(x)$,we can say that:

The equation of a tangent to the curve $y \cot x = y^3 \tan x$ at the point where the abscissa is $\frac{\pi}{4}$ is:

If $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$,where $f^{\prime \prime}(x) = -f(x)$ and $g(x) = f^{\prime}(x)$,and given $F(5) = 5$,then $F(10)$ is equal to

Let $k$ and $m$ be positive real numbers such that the function $f(x) = \begin{cases} 3x^2 + k\sqrt{x+1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}$ is differentiable for all $x > 0$. Then $\frac{8f'(8)}{f'(\frac{1}{8})}$ is equal to $.............$.

Let $f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 12$. Then