<s>[INST] Given that $x = 4\sin \left( {2y + 6} \right)$ find dy/dx in terms of x My attempt:
$\eqalign{
  & x = 4\sin \left( {2y + 6} \right)  \cr 
  & {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right)  \cr 
  & {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right)  \cr 
  & {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr} $
$\eqalign{
  & x = 4\sin \left( {2y + 6} \right)  \cr 
  & {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right)  \cr 
  & {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right)  \cr 
  & {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}}  \cr 
  & {\cos ^2}\left( {2y + 6} \right) + {\sin ^2}\left( {2y + 6} \right) = 1  \cr 
  & {\cos ^2}\left( {2y + 6} \right) + {\left( {{x \over 4}} \right)^2} = 1  \cr 
  & {\cos ^2}\left( {2y + 6} \right) = 1 - {{{x^2}} \over {16}}  \cr 
  & \cos \left( {2y + 6} \right) = \sqrt {{{16 - {x^2}} \over {16}}}   \cr 
  & \cos \left( {2y + 6} \right) = {{\sqrt {16 - {x^2}} } \over 4}  \cr 
  & {{dy} \over {dx}} = {1 \over {2\sqrt {16 - {x^2}} }} = 1 \cr} $
Okay I've got it right, but the official answer confuses me, it says:
${{dy} \over {dx}} = {1 \over {8cos\left( {\arcsin \left( {{x \over 4}} \right)} \right)}} = \left( {\left(  \pm  \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
This is the part i'm struggling to get my head around, although I arrive at the same answer. 
Okay 
$\arcsin {x \over 4} = 2y + 6$
but how does the answer then go : 
$\left( {\left(  \pm  \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
is there a shortcut or trick I overlooked?
I think I need some sleep, thanks...
 [/INST] You are on the right track.
Notice that 
$$\cos^2(2y+6) = 1-\frac{x^2}{16}$$ 
implies 
$$\cos(2y+6) = \pm\sqrt{1-\frac{x^2}{16}}.$$
</s>