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

##### 2 Answers

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

Let

where

Then

By the chain rule,

The product rule for three terms states:

If

Then

So,

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

#### Answer:

# 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

# 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) #