Question

Find `"dy"/"dx"`, if : x = `(t + 1/t)^a, y = a^(t+1/t)`, where a > 0, a ≠ 1, t ≠ 0.

Answer 1

x = `(t + 1/t), y = a(t + 1/t)`            ...(1)
Differentiating x and y w.r.t. x, we get
`"dx"/"dt" = "d"/"dt"(t + 1/t)^a`

= `a(t + 1/t)^(a - 1)."d"/"dt"(t + 1/t)`

= `a(t + 1/t)^(a - 1).(1 - 1/t^2)`
and
`"dy"/"dt" = "d"/"dt"[a^((t + 1/t))]`

= `a^((t + 1/t)).loga."d"/"dt"(t + 1/t)`

= `a^((t + 1/t)).loga.(1 - 1/t^2)`

∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt"))`

= `(a^((t + 1/t)).loga.(1 - 1/t^2))/(a(t + 1/t)^(a- 1).(1 - 1/t^2)`

= `(a^(t +1/t).loga.(t + 1/t))/(a.(t + 1/t)^a`

= `(yloga.((t^2 + 1)/t))/"ax"`         ...[By (1)]

= `(y(t^2 + 1)loga)/"axt"`.