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Question
Differentiate `tan^-1[(sqrt(1 + x^2) - 1)/x]` w.r. to `tan^-1[(2x sqrt(1 - x^2))/(1 - 2x^2)]`
Solution
Let u = `tan^-1[(sqrt(1 + x^2) - 1)/x]`
Put x = tan θ, then θ = tan–1x
∴ u = `tan^-1[(sqrt(1 + tan^2theta) - 1)/(tantheta)]`
= `tan^-1[(sqrt(sec^2theta) - 1)/(tantheta)]`
= `tan^-1[(sectheta - 1)/(tantheta)]`
= `tan^-1[(1/costheta - 1)/((sintheta)/(costheta))]`
= `tan^-1[(1 - costheta)/(sintheta)]`
= `tan^-1[(2sin^2(theta/2))/(2sin(theta/2)cos(theta/2))]`
= `tan^-1[tan theta/2]`
= `theta/2`
∴ u = `1/2 tan^-1x`
Differentiating w.r.t. x, we get
`("d"u)/("d"x) = 1/((2(1 + x^2)`
Let v = `tan^-1[(2xsqrt(1 - x^2))/(1 - 2x^2)]`
Put x = sin θ, then θ = sin–1x
∴ v = `tan^-1[(2sinthetasqrt(1 - sin^2theta))/(1 - 2sin^2theta)]`
= `tan^-1[(2sinthetacostheta)/(cos2theta)]`
= `tan^-1[(sin2theta)/(cos2theta)]`
= tan−1 (tan 2θ) = 2θ
∴ v = 2sin−1x
Differentiating w.r.t. x, we get
`("d"v)/("d"x) = 2/sqrt(1 - x^2)`
∴ `("d"u)/("d"x) = ((("d"u)/("d"x)))/((("d"v)/("d"x))`
= `1/(2(1 + x^2)) xx sqrt(1 - x^2)/2`
= `sqrt(1 - x^2)/(4(1 + x^2))`
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