Differential equation?

y=arc(cosx) these days the term arc( ) is not used! It is replaced by cos^(-1)x which does not mean it is 1/cosx , then it would be denoted as (cosx)^(-1). With this notation (cos^(-1)x)^(-1) would be
1/cos^(-1)x or in the old notation 1/arccos(x)

here we have y=cos^(-1)x (in words it means the angle whose cosine is y) x=cosy

dy/dx=- 1/siny but since x=cosy siny=sqrt(1-x^2) and substituting we obtain

dy/dx=- 1/sqrt(1-x^2)

y = arccos(x)

Use the trig functions that are easier to differentiate.
cos(y) = x

Use implicit differentiation here, in order to find dy/dx.
-sin(y) dy/dx = 1

Isolate:
dy/dx = -1/sin(y)

Replace y with above:
dy/dx = -1/sin(arccos(x))

Use the Pythagorean Identity to rewrite sin(w) as √[1 - cos²(w)]:
dy/dx = -1/√[1 - cos²(arccos(x))]

Simplify:
dy/dx = -1/√[1 - x²]