How do you use the Product Rule to find the derivative of #y=(x^2+1) (x+2)^2 (x-3)^3#?

1 Answer
Aug 17, 2015

Answer:

#color(red)( dy/dx =(x+2)(x-3)^2[3(x^2+1)(x+2) +2(2x^2+3)])#

Explanation:

#y= (x^2+1)(x+2)^2(x-3)^3#

The Product Rule, states that if #y=uvw#, then

#dy/dx = uv(dw)/dx+uw(dv)/dx + vw(du/dx)#

Let #u=x^2+1#, #v=(x+2)^2#, and #w = (x-3)^3#

Then

#(du)/dx=2x#, #(dv)/dx=2(x+2)#, and#(dw)/dx= 3(x-3)^2#

#dy/dx =uv(dw)/dx+uw(dv)/dx + vw(du/dx)#

#dy/dx=3(x^2+1)(x+2)^2(x-3)^2 + 2(x^2+1) (x+2) (x-3)^3+2x(x+2)^2(x-3)^2#

#dy/dx =(x+2)(x-3)^2[3(x^2+1)(x+2) +2(x^2+1)+2x(x+2)]#

#dy/dx =(x+2)(x-3)^2[3(x^2+1)(x+2) +2(2x^2+3)]#