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

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 #(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) #