Question

If y^x = e^{y – x}, prove that `"dy"/"dx" = (1 + log y)^2/logy`

Answer 1

Given that: y^x = e^{y – x} 

Taking log on both sides log y^x = log e^{y – x} 

⇒ x log y = (y – x)log e

⇒ x log y = y – x   .....[∵ log e = 1]

⇒ x log y + x = y

⇒ x(log y + 1) = y

⇒ x = `y/(log y + 1)`

Differentiating both sides w.r.t. y

`"dx"/"dy" = "d"/"dy"(y/(log y + 1))`

= `((log y + 1) * 1 - y * "d"/"dy" (log y + 1))/(log y + 1)^2`

= `(log y + 1 - y * 1/2)/(log y + 1)^2`

= `logy/(log y + 1)^2`

We know that

`"dy"/"dx" = 1/("dx"/"dy")`

= `1/(logy/(log y + 1)^2`

= `(log y + 1)^2/logy`

Hence, `"dy"/"dx" = (log y + 1)^2/logy`.