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:
# 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) #