But i need a more detail explanation find the derivative function of the differential calculus (2x+3)^3(4x^2-1)^2

To find the derivative function of the given expression, (2x+3)^3(4x^2-1)^2, we can apply the product rule and the chain rule of differentiation.

Let's break down the expression into two parts:
Part 1: (2x+3)^3
Part 2: (4x^2-1)^2

First, let's find the derivative of Part 1 using the chain rule:
Let u = 2x + 3
Therefore, Part 1 = u^3, and we need to find du/dx.

To find du/dx, take the derivative of u with respect to x:
du/dx = d/dx (2x + 3)
= 2

Now, apply the chain rule:
d(u^3)/dx = 3u^2 * du/dx
= 3(2x + 3)^2 * 2

Simplifying, we get:
d(u^3)/dx = 6(2x + 3)^2

Next, let's find the derivative of Part 2:
d(4x^2-1)^2/dx

Using the chain rule, let's set v = 4x^2 - 1. Therefore, Part 2 = v^2, and we need to find dv/dx.

To find dv/dx, take the derivative of v with respect to x:
dv/dx = d/dx (4x^2 - 1)
= 8x

Now, apply the chain rule:
d(v^2)/dx = 2v * dv/dx
= 2(4x^2 - 1) * 8x
= 16x(4x^2 - 1)

Now, apply the product rule to find the derivative of the entire expression:
d/dx [(2x+3)^3(4x^2-1)^2] = (2x+3)^3 * d[(4x^2-1)^2]/dx + (4x^2-1)^2 * d[(2x+3)^3]/dx

Using the derivatives we previously found:
= (2x+3)^3 * 16x(4x^2 - 1) + (4x^2-1)^2 * 6(2x + 3)^2

Simplifying, we can factor out a common term of (2x+3)^2:
d/dx [(2x+3)^3(4x^2-1)^2] = (2x+3)^2 * [16x(4x^2 - 1) + 6(4x^2-1)]

Further simplification gives the final derivative function:
d/dx [(2x+3)^3(4x^2-1)^2] = (2x+3)^2 * [64x^3 - 16x + 24x^2 - 6]

To find the derivative of the given function (2x+3)^3(4x^2-1)^2, we can use the product rule and the chain rule.

The product rule states that if we have two functions, u(x) and v(x), then the derivative of their product is given by:

(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)

Let's denote the first part of the product, (2x+3)^3, as u(x), and the second part of the product, (4x^2-1)^2, as v(x).

First, let's find the derivative of u(x):

u'(x) = d/dx((2x+3)^3)

To compute the derivative here, we will use the chain rule, which states that if we have a composite function f(g(x)), then its derivative is given by:

(d/dx)(f(g(x))) = f'(g(x)) * g'(x)

In our case, f(x) = x^3, and g(x) = 2x+3. Therefore, we have:

u'(x) = d/dx((2x+3)^3)
= 3(2x+3)^2 * d/dx(2x+3)
= 3(2x+3)^2 * 2

Simplifying further, we get:

u'(x) = 6(2x+3)^2

Now let's find the derivative of v(x):

v'(x) = d/dx((4x^2-1)^2)

Again, using the chain rule, we set f(x) = x^2 and g(x) = 4x^2-1:

v'(x) = d/dx((4x^2-1)^2)
= 2(4x^2-1) * d/dx(4x^2-1)
= 2(4x^2-1) * d/dx(4x^2) (The derivative of -1 is 0)
= 2(4x^2-1) * 8x
= 16x(4x^2-1)

Now, using the product rule, we can find the derivative of the original function:

(d/dx)((2x+3)^3(4x^2-1)^2)
= u'(x)v(x) + u(x)v'(x)
= (6(2x+3)^2)(4x^2-1)^2 + (2x+3)^3(16x(4x^2-1))

Expanding and simplifying this expression may be further required depending on the context of the problem.