Question

Find the derivative of cos(2tan^-1((1-x)÷(1+x))^(1÷2))

Answer 1

By definition, tanθ=sinθcosθ=sinθ√1−sin2θ, so

tan2θ(1−sin2θ)=sin2θ

tan2θ=sin2θ(1+tan2θ)

sin2θ=tan2θ1+tan2θ

By definition, tanarctanx=x, so 1−2sin2arctanx becomes 1−2x21+x2. Putting this over a common denominator makes 1−x21+x2.

So

cos(2arctanx)=1−x21+x2.

Answer 2

definition, tanθ=sinθcosθ=sinθ√1−sin2θ, so

tan2θ(1−sin2θ)=sin2θ

tan2θ=sin2θ(1+tan2θ)

sin2θ=tan2θ1+tan2θ

By definition, tanarctanx=x, so 1−2sin2arctanx becomes 1−2x21+x2. Putting this over a common denominator makes 1−x21+x2.

So

cos(2arctanx)=1−x21+x2.