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

2 Answers
Sep 10, 2017

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

Sep 10, 2017

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)