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 :

Let x^k + y^k = a^k, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:

If $$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right)$$

$${{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is$$ :

The derivative of $${\tan ^{ - 1}}\left( {{{\sin x - \cos x} \over {\sin x + \cos x}}} \right)$$, with respect to $${x \over 2}$$ , where $$\left( {x \in \left( {0,{\pi \over 2}} \right)} \right)$$ is :