Question

If y = log `("e"^"x"/"x"^2)`, then `"dy"/"dx" = ?` 

Options

• `(2 - "x")/"x"`

• `("x" - 2)/"x"`

• `("e - x")/"ex"`

• `("x - e")/"ex"`

Answer 1

`bb(("x" - 2)/"x")`

Explanation:

y = log `("e"^"x"/"x"^2)`

= log (e^x) − log (x²)

= x log e − log x²

y = x − log x²   ...(∵ log e = 1)

Differentiating w.r.t. 'x', we get

`"dy"/"dx"= 1 - 1/("x"^2)."d"/"dx" ("x"^2)`

= `1 - (2"x")/("x"^2)`

= `1 - 2/"x"`

= `("x"- 2)/"x"`