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Question
Find the derivatives of the following:
If u = `tan^-1 (sqrt(1 + x^2) - 1)/x` and v = `tan^-1 x`, find `("d"u)/("d"v)`
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Solution
u = `tan^-1 (sqrt(1 + x^2) - 1)/x`
Put x = tan θ,
u = `tan^-1 ((sqrt(1 + tan^2theta) - 1)/tan theta)`
= `tan^-1 ((sqrt(sec^2theta) - 1)/tan theta)`
= `tan^-1 ((sectheta - 1)/tantheta)`
= `tan^-1 ((1/(costheta) - 1)/((sintheta)/costheta))`
= `tan^-1 (((1 - costheta)/costheta)/((sin theta)/(costheta)))`
= `tan^-1 ((1 - cos theta)/sintheta)`
= `tan^-1 ((2 sin^2 theta/2)/(2 sin theta/2 cos theta/2))`
= `tan^-1 ((sin theta/2)/(cos theta/2))`
= `tan^-1 (tan theta/2)`
= `theta/2`
u = `1/2 tan^-1 (x)`
`("d"u)/("d"x) = 1/2 xx 1/(1 + x^2)` .......(1)
Let v = `tan^-1 (x)`
`("d"v)/("d"x) = 1/(1 + x^2)` ........(2)
From equations (1) and (2)
`(("d"u)/("d"x))/(("d"v)/("d"x)) = (1/(2(1 + x^2)))/(1/(1 + x^2))`
`("d"u)/("d"v) = 1/2`
`("d"(tan^-1 ((sqrt(1 + x^2) - 1)/x)))/("d"(tan^-1 x)) = 1/2`
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