Find derivative: (write answer as single fraction)

y=ln((4x-1)/x^3)

So far I have:
y'=(4/(4x-1))-3/x
How do I write as single fraction

let u= (4x-1) v= x^-3

dy/dx= d y/d(uv) * d(uv)/dx

dy/duv= 1/uv

d(uv)/dx= udv/dx + v du/dx

put it together.

To write the derivative as a single fraction, we can simplify the expression by combining the terms involving u and v. Let's start with what you have:

First, we have dy/duv = 1/uv.

Next, let's calculate the partial derivatives du/dx and dv/dx:

To find du/dx, we differentiate u with respect to x, which is the same as finding the derivative of (4x - 1) with respect to x. Since the derivative of a constant is zero, the derivative of -1 with respect to x is 0. For the term 4x, the derivative with respect to x is simply 4. Therefore, du/dx = 4.

To find dv/dx, we differentiate v with respect to x, which is the same as finding the derivative of x^-3 with respect to x. Using the power rule for differentiation, we get dv/dx = -3x^-4 = -3/x^4.

Now, let's calculate udv/dx and v du/dx:

udv/dx = (4x - 1)(-3/x^4) = -12(4x - 1)/x^4.

v du/dx = (x^-3)(4) = 4/x^3.

Finally, we can combine the terms:

d(uv)/dx = udv/dx + v du/dx = -12(4x - 1)/x^4 + 4/x^3.

Therefore, the derivative dy/dx of y = ln((4x - 1)/x^3) written as a single fraction is:

dy/dx = (1/uv)(-12(4x - 1)/x^4 + 4/x^3).

Simplifying further, we have:

dy/dx = (-12(4x - 1))/x^5 + 4/x^4.

So, the derivative as a single fraction is:

dy/dx = (-48x + 12)/x^5 + 4/x^4.