Question

Choose the correct alternative.

If y = `sqrt("x" + 1/"x")`, then `"dy"/"dx" = ?`

Options

• `("x"^2 - 1)/(2"x"^2sqrt("x"^2 + 1))`

• `(1 - "x"^2)/(2"x"^2("x"^2 + 1))`

• `("x"^2 - 1)/("2x"sqrt"x"sqrt("x"^2 + 1))`

• `(1 - "x"^2)/("2x"sqrt"x"sqrt("x"^2 + 1))`

Answer 1

`("x"^2 - 1)/("2x"sqrt"x"sqrt("x"^2 + 1))`

Explanation:

y = `sqrt("x" + 1/"x")`

Differentiating both sides w.r.t.x, we get

`"dy"/"dx" = 1/(2sqrt("x" + 1/"x")) * "d"/"dx" ("x" + 1/"x")`

`= 1/(2sqrt(("x"^2 + 1)/"x")) * (1 - 1/"x"^2)`

`= sqrt"x"/(2sqrt("x"^2 + 1)) * (("x"^2 - 1)/"x"^2)`

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