Question

How do you differentiate `f(x) =(x^2+1) (x+2)^2 (x-3)^3` using the product rule?

Answer 1

First, split off your separate expressions into sub-functions.

Let `y=t*u*v`
where `t=x^2+1`, `u=(x+2)^2`, and `v=(x-3)^3`

Then `dt/dx = 2x`.

`(du)/dx = 2(x+2)`.

By the chain rule, `(dv)/dx = 3(x-3)^2`.

The product rule for three terms states:

If `y=t*u*v`, and `y` is a function of `x`.

Then `dy/dx = dt/dx*u*v + (du)/dx*t*v + (dv)/dx*t*u`.

So, `dy/dx =` `2x*(x+2)^2*(x-3)^3+2(x^2+1)*(x-3)^3*(x+2)+3(x^2 + 1)*(x+2)^2*(x-3)^3`

Which when you go through the painful process of expansion and simplification, yields:

`dy/dx=7x^6-30x^5-20x^4+160x^3-15x^2-126x`

Answer 2

`f'(x) = x(x+2)(x-3)^2 (7x^2-2x-7)`

Explanation:

We have:

`f(x) = (x^2+1)(x+2)^2(x-3)^3`

We can utilise the triple product ##rule, a direct extension of the standard product rule for differentiation:

`d/dx(uvw) = uv(dw)/dx+u(dv)/dxw + (du)/dxvw`

and we will also require the chain rule:

Applying the triple product## rule we get:

`f'(x) = (x^2+1)(x+2)^2(d/dx(x-3)^3) +`
`\ \ \ \ \ \ \ \ \ \ \ (x^2+1)(d/dx (x+2)^2)(x-3)^3 +`
`\ \ \ \ \ \ \ \ \ \ \ (d/dx (x^2+1))(x+2)^2(x-3)^3`

`\ \ \ \ \ \ \ \ \= (x^2+1)(x+2)^2(3(x-3)^2) +`
`\ \ \ \ \ \ \ \ \ \ \ (x^2+1)(2(x+2))(x-3)^3 +`
`\ \ \ \ \ \ \ \ \ \ \ (2x)(x+2)^2(x-3)^3`

We can readily simplify as there is a common factor of `(x+2)(x-3)^2`:

`f'(x) = (x+2)(x-3)^2 { 3(x^2+1)(x+2) +`
`\ \ \ \ \ \ \ \ \ \ \ 2(x^2+1)(x-3) +`
`\ \ \ \ \ \ \ \ \ \ \ (2x)(x+2)(x-3) }`

`\ \ \ \ \ \ \ \ \= (x+2)(x-3)^2 { (3x^3+6x^2+3x+6) +`
`\ \ \ \ \ \ \ \ \ \ \ (2x^3-6x^2+2x-6) + (2x^3-2x^2-12x) }`

`\ \ \ \ \ \ \ \ \= (x+2)(x-3)^2 (7x^3-2x^2-7x)`
`\ \ \ \ \ \ \ \ \= x(x+2)(x-3)^2 (7x^2-2x-7)`