Question

Derivative of `tan^-1(x/sqrt(1 - x^2))` with respect sin^–1(3x – 4x³) is ______.

Options

• `1/sqrt(1 - x^2)`

• `3/sqrt(1 - x^2)`

• 3

• `1/3`

Answer 1

Derivative of `tan^-1(x/sqrt(1 - x^2))` with respect sin^–1(3x – 4x³) is `underlinebb(1/3)`.

Explanation:

Let u = `tan^-1(x/sqrt(1 - x^2))`

and v = sin^–1 (3x – 4x³)

Now, put x = sin θ `\implies` θ = sin^–1(x), then

u = `tan^-1((sinθ)/sqrt(1 - sin^2θ))`

and v = sin^–1 (3 sin θ – 4 sin³ θ)

`\implies` u = `tan^-1(sinθ/cosθ)` and v = sin^–1 (sin 3 θ)

`\implies` u = tan^–1 (tan θ) and v = sin^–1 (sin 3 θ)

`\implies` u = θ and v = 3 θ

`\implies` u = sin^–1x and v = 3sin^–1x.

On differentiating both sides w.r.t. x, we get

`(du)/dx = 1/sqrt(1 - x^2)` and `(dv)/dx = 3 xx 1/sqrt(1 - x^2)`

∴ `(du)/(dv) = ((du)/dx)/((dv)/dx) = (1/sqrt(1 - x^2))/(3/sqrt(1 - x^2)) = 1/3`