Question

If for x ∈ `(0, 1/4)`, the derivative of `tan^-1((6xsqrt(x))/(1 - 9x^3))` is `sqrt(x) * "g"(x)`, then g(x) equals ______.

Options

• `3/(1 + 9x^3)`

• `9/(1 + 9x^3)`

• `(3xsqrt(x))/(1 - 9x^3)`

• `(x)/(1 - 9x^3)`

Answer 1

If for x ∈ `(0, 1/4)`, the derivative of `tan^-1((6xsqrt(x))/(1 - 9x^3))` is `sqrt(x) * "g"(x)`, then g(x) equals `9/(1 + 9x^3)`.

Explanation:

Let y = `tan^-1((6xsqrt(x))/(1 - 9x^3))`

= `tan^-1 [(6x^(3/2))/(1 - (3x^(3/2))^2)]`

= `tan^-1 [(2 xx 3x^(3/2))/(1 - (3x^(3/2))^2)]`

= `2tan^-1 3x^(3/2)`

∴ `("d"y)/("d"x) = 2/(1 + (3x^(3/2))^2) * 3 xx 3/2 xx x^(1/2)`

= `9/(1 + 9x^3)sqrt(x)`

Comparing with `sqrt(x)  "g"(x)`, we get

g(x) = `9/(1 + 9x^3)`