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Question
x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`
Solution
Given that: x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`
Differentiating both the parametric functions w.r.t. t
`"dx"/"dt" = ("t"^2 * "d"/"dt" (1 + log "t") - (1 + log "t") * "d"/"dt" ("t"^2))/"t"^4`
= `("t"^2 * (1/"t") - (1 + log "t") * 2"t")/"t"^4`
= `("t" - (1 + log "t") * 2"t")/"t"^4`
= `("t"[1 - 2 - 2 log "t"])/"t"^4`
= `(-(1 + 2 log "t"))/"t"^3`
y = `(3 + 2 log "t")/"t"`
`"dy"/"dt" = ("t" * "d"/"dt" (3 + 2 log "t") - (3 + 2 log "t") * "d"/"dt" ("t"))/"t"^2`
= `("t"(2/"t") - (3 + 2 log "t")* 1)/"t"^2`
= `(2 - 3 - 2 log "t")/"t"^2`
= `(-(1 + 2 log "t"))/"t"^2`
∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`
= `((-(1 + 2 log "t"))/"t"^2)/((-(1 + 2 log "t"))/"t"^3)`
= `"t"^3/"t"^2`
= t
Hence, `"dy"/"dx"` = t.
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