How do you differentiate y=arccos(1-2x^2) with respect to x, and simplifying the answers

let 1 - 2x^2 = t

-4x = dt/dx

y = arccos(1 - 2x^2)

=> y = arccos(t)

dy/dx = -1/√(1 - t^2)* dt/dx

substitute t = 1 - 2x^2 and dt/dx = -4x

dy/dx = 4x/√[1 - (1 - 2x^2)^2 ]

= 4x /√(1 - 1 - 4x^4 + 4x^2)

= 4x/√(4x^2 - 4x^4)

= 4x /2x√(1 - x^2)

= 2/√(1 - x^2)

First, I would rewrite the relation as follows:

cos y = 1 - 2x^2

Next, d(cos y)/dx = - sin y dy/dx and

d(1 - 2x^2)/dx = - 4x

So, - sin y dy/dx = - 4x

Solving for dy/dx: dy/dx = 4x / sin y

Since cos y = 1 - 2x^2, and sin^2 y + cos^2 y = 1, sin y = ± (1 - [1 - 2x^2]^2)^(1/2)

Simplifying the last expression:

(1 - [1 - 2x^2]^2)^(1/2) = (1 - [1 - 4x^2 + 4x^4])^(1/2) = 4x^2 (1 - x^2)

Then, dy/dx = ± 4x / [4x^2 (1 - x^2)] = ± 1/ (x [1 - x^2])

... y = arccos(1 - 2x²)
or cos(y) = 1 - 2x²
or - sin(y) dy/dx = - 4x
or dy/dx = 4x / sin(y)
or dy/dx = 4x / √ ( 1 - cos²(y) )
or dy/dx = 4x / √ ( 1 - (1 - 2x²)² )
or dy/dx = 4x / [ 2x √ (1 - x²) ]
or dy/dx = 2 / √ (1 - x²)