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Question
If x = log(1 + t2), y = t – tan–1t,show that `"dy"/"dx" = sqrt(e^x - 1)/(2)`.
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Solution
x = log(1 + t2), y = t – tan–1t
Differentiating x and y w.r.t. t, we get
`"dx"/"dt" = "d"/"dt"[log(1 + t^2)]`
= `(1)/(1 + t^2)."d"/"dt"(1 - t^2)`
= `(1)/(1 + t^2) xx (0 + 2t)`
= `(2t)/(1 + t^2)`
and
`"dy"/"dt" = "d"/"dt"(t) - "d"/"dt"(tan^-1t)`
= `1 - (1)/(1 + t^2)`
= `(1 + t^2 - 1)/(1 + t^2)`
= `t^2/(1 + t2)`
∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`
= `(((t2)/(1 + t^2)))/(((2t)/(1 + t^2))`
= `t/(2)`
Now, x = log (1 + t2)
∴ 1 + t2 = ex
∴ t2 = ex - 1
∴ t = `sqrt(e^x - 1)`
∴ `"dy"/"dx" = sqrt(e^x - 1)/(2)`.
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