y= arcos(x). Find dy/dx in terms of x.

Rearrange the expression to create a familiar function with a known differencial: arcos(x)=y x= cos(y) Differenciate x with respect to y: dx/dy= -sin(y) We know that dy/dx= 1/(dx/dy), so rearange to find an expression for 'dy/dx' in terms of 'y': dy/dx= -1/sin(y) The answer asks for 'dy/dx' in terms of 'x', so we need to find an equation linking 'y' to 'x'; we already know that 'x=cos(y'). We now need an equation linking 'sin(y') to 'cos(y)'; we can use the trigonometric identity: 'sin^2(y)+ cos^2(y)= 1' or 'sin(y)= sqr root(1- cos^2(y))' If we subsitute sin(y) for sqr root(1- cos^2(y)) and 'x' for 'cos(y)' We arrive at the answer: dy/dx= -1/{ sqr root[1- cos^2(y)]}