# Find dydx if, xy=ex - y - Mathematics and Statistics

Sum

Find "dy"/"dx" if, "x"^"y" = "e"^("x - y")

#### Solution

"x"^"y" = "e"^("x - y")

Taking logarithm of both sides, we get

y log x = (x - y) log e = x - y

∴ y log x + y = x

∴ y(1 + log x) = x

∴ y = "x"/(1 + log "x")

Differentiating both sides w.r.t. x, we get

"dy"/"dx" = "d"/"dx" ["x"/(1 + log "x")]

∴ "dy"/"dx" = ((1 + log "x") "d"/"dx" ("x") - "x" "d"/"dx" (1 + log "x"))/(1 + log "x")^2

= ((1 + log "x") xx 1 - "x" xx (1/"x"))/(1 + log "x")^2

= (1 + log "x" - 1)/(1 + log "x")^2

∴ "dy"/"dx" = (log "x")/(1 + log "x")^2

Concept: Derivatives of Implicit Functions
Is there an error in this question or solution?