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Question:

Differentiate tan⁻¹(√(1-x²)/x) with respect to sin⁻¹(2x/(1+x²)), when x≠0.

Solution:

y=tan⁻¹(√(1+x²)/x)
y=tan⁻¹(√(1+tan²θ)/tanθ)
Put x=tanθ
tan⁻¹(secθ/tanθ)=tan⁻¹((1/cosθ)/(sinθ/cosθ))
=tan⁻¹(1/sinθ)
=tan⁻¹(cosecθ)=tan⁻¹(2sin(θ/2)cos(θ/2)/2sin(θ/2)cos(θ/2))
=tan⁻¹(1/(sinθ))
=tan⁻¹(1/(2sin(θ/2)cos(θ/2)))
=tan⁻¹(1/sinθ) = θ/2
dy/dθ=1/2
Let z=sin⁻¹(2x/(1+x²))=sin⁻¹(2tanθ/(1+tan²θ))
=sin⁻¹(sin2θ)=2θ
dz/dθ=2
Now, dy/dz=(dy/dθ)(dθ/dz)=(1/2)(1/2)=1/4