Question

If `y = (x + sqrt(1 + x^2))^n`, then `(1 + x^2) (d^2y)/(dx^2) + x (dy)/(dx)` is

Options

• n²y

• – n²y

• – y

• 2x²y

Answer 1

n²y

Explanation:

`y = (x + sqrt(1 + x^2))^n`

Differentiate w.r.t.x

`(dy)/(dx) = n(x + sqrt(1 + x^2))^(n-1) [(1 + 1)/(2sqrt(1 + x^2) × 2x)]`

= `n(x + sqrt(1 + x^2))^(n-1) [(1 + x)/sqrt(1 + x^2)]`

= `n(x + sqrt(1 + x^2))^(n-1) [(sqrt(1 + x^2) + x)/sqrt(1 + x^2)]`

= `(n(x + sqrt(1 + x^2))^n)/sqrt(1 + x^2)`

= `(ny)/sqrt(1 + x^2)`

`sqrt(1 + x^2) (dy)/(dx)` = ny  ......(i)

Differentiate again w.r.t.x

`[(2x)/2sqrt(1 + x^2)] × (dy)/(dx) + sqrt(1 + x^2) (d^2y)/(dx^2) = n (dy)/(dx) [x/sqrt(1 + x^2)] xx (dy)/(dx) + sqrt(1 + x^2) (d^2y)/(dx^2) = n (dy)/(dx)`

Multiply by `sqrt(1 + x^2)`

`[x xx (dy)/(dx) + (1 + x^2) (d^2y)/(dx^2)] = n ((dy)/(dx)) sqrt(1 + x^2) [x xx (dy)/(dx) + (1 + x^2) (d^2y)/(dx^2)]` = n ny  ......(From (i))

`(1 + x^2) (d^2y)/(dx^2) + x (dy)/(dx)` = n²y