Question

If x = log(1 + t²), y = t – tan^–1t,show that `"dy"/"dx" = sqrt(e^x - 1)/(2)`.

Answer 1

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 = e^x
∴  t² = e^x - 1
∴  t = `sqrt(e^x - 1)`
∴  `"dy"/"dx" = sqrt(e^x - 1)/(2)`.