Differentiation

Let y = y(x) be a function of x satisfying $y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}$ where k is a constant and $y\left( {{1 \over 2}} \right) = - {1 \over 4}$. Then ${{dy} \over {dx}}$ at x = ${1 \over 2}$, is equal to :

If $\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$ where a > b > 0, then ${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$ is :

Let $y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)$. Then, at x = 1,

Let $f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}$. Then $f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)$ is equal to

If $2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$, x $\in$ $\left( {0,{\pi \over 2}} \right)$ then $dy \over dx$ is equal to:

Let ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1, |x| > 1. If ${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$ and $y\left( {\sqrt 3 } \right) = {\pi \over 6}$, then y(${ - \sqrt 3 }$) is equal to :

Let $f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$, 0 < x < 1. Then :

If $y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$, then ${{dy} \over {dx}}$ at $x = {{5\pi } \over 6}$ is :

If $y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$, then

If $\log _e y=3 \sin ^{-1} x$, then $(1-x^2) y^{\prime \prime}-x y^{\prime}$ at $x=\frac{1}{2}$ is equal to