If for x∈(0,1/4), the derivative of tan⁻¹(6x√x+19x³) is √x⋅g(x), then g(x) equals.

Let y = tan⁻¹(6x√x + 19x³)
Then, dy/dx = 1/(1+(6x√x + 19x³)² ) * d(6x√x + 19x³)/dx
= 1/(1+(6x√x + 19x³)² ) * (6(x^(3/2)) + 57x²)
Given that dy/dx = √x * g(x)
Therefore, g(x) = (1/(1+(6x√x + 19x³)² ) * (6(x^(3/2)) + 57x²))/√x
Let u = 6x√x + 19x³ = 6x^(3/2) + 19x³
Then du/dx = 9x^(1/2) + 57x²
Let y = tan⁻¹(u)
Then dy/du = 1/(1+u²)
dy/dx = dy/du * du/dx = (1/(1+u²)) * (9√x + 57x²)
dy/dx = √x * g(x)
g(x) = (1/(1+u²)) * (9 + 57x^(3/2))/√x
If we let u = 3x√x = 3x^(3/2), then u² = 9x³
du/dx = (9/2)√x
dy/dx = 1/(1+9x³) * (9/2)√x
Let's consider the function: y = tan⁻¹(3x√x)
Then dy/dx = [1/(1+(3x√x)²)] * d(3x^(3/2))/dx
= 1/(1+9x³) * (9/2)√x
Therefore, √x * g(x) = (9/2)√x/(1+9x³)
g(x) = 9/(2(1+9x³))
This doesn't match any of the options.
Let's reconsider the derivative:
Let u = 6x√x + 19x³
Then du/dx = 9x^(1/2) + 57x² = 9√x + 57x²
dy/dx = 1/(1+u²) * (9√x + 57x²) = √x * g(x)
g(x) = (9 + 57x^(3/2))/(√x(1+(6x√x + 19x³)²))
Let's assume the question meant tan⁻¹(3x√x)
Then dy/dx = 1/(1+(3x√x)²) * (9/2)√x
= (9/2)√x/(1+9x³)
Then g(x) = 9/(2(1+9x³))
This is still not one of the options. There seems to be an error in the problem statement or options.