# If Xy - Yx = Ab, Find D Y D X . - Mathematics

Sum

If xy - yx = ab, find (dy)/(dx).

#### Solution

The given function is xy - yx = ab
Let xy = u and yx = v
Then , the function becomes u - v = ab

(d"u")/(d"x") - (d"v")/(d"x") = 0         .....(1)
u = xy
⇒ log u = log(xy)
⇒ log u = y log x

Differentiating both sides with respect to x, we obtain

(1)/("u") (d"u")/(d"x") = log "x" (d"y")/(d"x") + "y". (d)/(d"x") (log "x")

⇒ (d"u")/(d"x") = [ log "x" (d"y")/(d"x") + "y" .(1)/("x")]

⇒ (d"u")/(d"x") = "x"^"y" (log "x"  (d"y")/(d"x") + ("y")/("x"))     ...(2)

"v" = "y"^"x"

⇒ log "v" = log ("y"^"x")

⇒ log "v" = "x" log "y"

Differentiating both sides with respect to x, we obtain

(1)/("v").(d"v")/(d"x") = log "y".(d)/(d"x") ("x") + "x". (d)/(d"x") (log "y")

⇒ (d"v")/(d"x") = "v" (log "y". 1 + "x".(1)/("y"). (d"y")/(d"x"))

⇒ (d"v")/(d"x") = "y"^"x" (log "y" + ("x")/("y") (d"y")/(d"x"))      ...(3)

From (1), (2), and (3), we obtain

"x"^"y" (log "x"(d"y")/(d"x") + ("y")/("x")) - "y"^"x" (log "y" + ("x")/("y") (d"y")/(d"x")) = 0

⇒ "x"^"y" log "x" (d"y")/(d"x") - "xy"^("x"-1) (d"y")/(d"x") + "x"^("y"-1) "y"-"y"^"x" log"y" = 0

⇒ ("x"^"y" log "x" - "xy"^("x"-1)) (d"y")/(d"x") = "y"^"x" log "y" - "x"^("y"-1) "y"

⇒ (d"y")/(d"x") = ("y"^"x" log "y" - "x"^("y" -1) "y")/(("x"^"y" log "x" - "xy"^("x"-1))

Concept: Exponential and Logarithmic Functions
Is there an error in this question or solution?