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