Find dy/dx if y = tan¯¹(2x/1-x²)
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Given y = tan–1 (2x/ 1 - x2) Put x = tan θ ⇒ θ = tan–1x y = tan–1 (2 tan θ/ 1 - tan2 θ) = tan–1 (tan 2θ) = 2θ = 2tan–1x differentiating w.r.to 'x' on both sides, we have (dy/dx) = 2(d/dx) = tan−1 x = 2(1/1 + x2) ∴ (dy/dx) = (2/ 1 + x2)
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