Question

Differentiate tan−1(2x1−x2) with respect to sin−1(2x1+x2)?

Answer 1

Dear Student,

Let u=tan−1(1+x2√−1x) and v=tan−1(x)Put x = tan θNow, u = tan−1[1 + tan2θ√ − 1tan θ]⇒u = tan−1[sec θ − 1tan θ]⇒u = tan−1[1 − cos θsin θ]⇒u = tan−1[2 sin2(θ/2)2 sin(θ/2) . cos(θ/2)]⇒u = tan−1[tan(θ/2)] = θ2 = 12tan−1x⇒dudx = 12.11+x2Now, v =tan−1x⇒v = tan−1[tan θ]⇒v = θ⇒v =tan−1x⇒dvdx = 11+x2Now, dudv = dudx×dxdv = 12.11+x2 × 1+x21 = 12

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