The question is asking to prove that:
d(arccos(x))/dx≡-1/√(1-x²) which can be written:
arccos(x)≡∫(-1/√(1-x²))dx.
Let x=cosθ, then dx=-sinθdθ=-√(1-x²)dθ, so dθ=-dx/√(1-x²).
Hence ∫(-1/√(1-x²))dx=∫dθ=θ+c where c is constant of integration.
Since θ=arccos(x) because x=cosθ, d(arccos(x))/dx=dθ/dx=-1/√(1-x²) QED