Question

If x^y = e^x-y, then `"dy"/"dx"` at x = 1 is ______.

Options

• e

• 1

• 0

• -1

Answer 1

If x^y = e^x-y, then `"dy"/"dx"` at x = 1 is 0.

Explanation:

Given have, x^y = e^x-y 

taking log on both sides, we get

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

when x = 1, then y (log 1) = (1 - y)

⇒ y = 1

On differentiating both sides,

`"y"(1/x) + log x * "dy"/"dx" = 1 - "dy"/"dx"`

`=> "dy"/"dx" (log x + 1) = 1 - "y"/x`

`=> "dy"/"dx" (log x + 1) = (x - y)/x`

`=> "dy"/"dx" = (x - y)/(x(log x + 1))`

when x = 1, then

`("dy"/"dx") = (1 - 1)/(1 (log 1 + 1))` = 0