$$f(x)$$ is a differentiable function and $$g(x)$$ is a double differentiable
function such that $$\left| {f\left( x \right)} \right| \le 1$$ and $$f'(x)=g(x).$$
If $${f^2}\left( 0 \right) + {g^2}\left( 0 \right) = 9.$$ Prove that there exists some $$c \in \left( { - 3,3} \right)$$
such that $$g(c).g''(c)<0.$$

If$$\,\,\,$$ $$y = {{a{x^2}} \over {\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + {{bx} \over {\left( {x - b} \right)\left( {x - c} \right)}} + {c \over {x - c}} + 1$$,
prove that $${{y'} \over y} = {1 \over x}\left( {{a \over {a - x}} + {b \over {b - x}} + {c \over {c - x}}} \right)$$.

Find $${{{dy} \over {dx}}}$$ at $$x=-1$$, when
$${\left( {\sin y} \right)^{\sin \left( {{\pi \over 2}x} \right)}} + {{\sqrt 3 } \over 2}{\sec ^{ - 1}}\left( {2x} \right) + {2^x}\tan \left( {In\left( {x + 2} \right)} \right) = 0$$

If $$x = \sec \theta - \cos \theta $$ and $$y = {\sec ^n}\theta - {\cos ^n}\theta $$, then show
that $$\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$$