Problem: Find dy/dx of the given function:
Y=1/2cos-1(1-4x3)/(1+4x3)

Solution:

We can use the chain rule to find the derivative of the given function.

Step 1: Rewrite the function as:
y = (1/2)cos^-1[(1-4x^3)/(1+4x^3)]

Step 2: Use the chain rule to find dy/dx:
dy/dx = (dy/du) * (du/dx)

where u = (1-4x^3)/(1+4x^3)

Step 3: Find du/dx:
du/dx = [(1+4x^3)(-12x^2)-(1-4x^3)(12x^2)]/(1+4x^3)^2

du/dx = -96x^5/(1+4x^3)^2

Step 4: Find dy/du:
dy/du = -sin^-1(u)'

where u = (1-4x^3)/(1+4x^3)

dy/du = -1/(sqrt(1-u^2))

dy/du = -1/(sqrt[1-((1-4x^3)/(1+4x^3))^2])

dy/du = -1/(sqrt[1-((1-4x^3)^2/(1+4x^3)^2)])

dy/du = -1/(sqrt[(8x^3+1)/(8x^3+1)^2)])

dy/du = -sqrt(8x^3+1)/(8x^3+1)

Step 5: Substitute du/dx and dy/du in the chain rule formula:
dy/dx = (dy/du) * (du/dx)

dy/dx = [-sqrt(8x^3+1)/(8x^3+1)] * [-96x^5/(1+4x^3)^2]

dy/dx = (96x^5)/(1+4x^3)^2 * (sqrt(8x^3+1)/(8x^3+1))

Therefore, the derivative of the given function is:
dy/dx = (96x^5 * sqrt(8x^3+1))/(1+4x^3)^2(8x^3+1)

Hence, the final solution is:
dy/dx = (96x^5 * sqrt(8x^3+1))/(1+4x^3)^2(8x^3+1)