Question

The derivative oftan−1⁡(1+x2−1x) with respect to tan−1⁡(2x1−x21−2x2) at x=0

• A. (A) 18
• B. (B) 14
• C. (C) 12
• D. (D) 6

Answer 1

Answer: (B)

(B) 14

Answer 2

Explanation:
Let y=tan−1⁡(1+x2−1x) and z=tan−1⁡(2x1−x21−2x2)Put x=tan⁡θ in y, we gety=tan−1⁡(sec2⁡θ−1tan⁡θ)=tan−1⁡(sec⁡θ−1tan⁡θ)=tan−1⁡(tan⁡θ2)=12tan−1⁡x⇒dydx=12(1+x2)Put x=sin⁡θ in z, we getz=tan−1⁡(2sin⁡θcos2⁡θ1−2sin2⁡θ)=tan−1⁡(2sin⁡θcos⁡θcos⁡2θ)=tan−1⁡(tan⁡2θ)=2θz=2sin−1⁡x⇒dzdx=21−x2 Thus, dydz=dy/dxdz/dx=12(1+x2)×1−x22Atx=0,dydz=14