How do you apply the product rule repeatedly to find the derivative of #f(x) = (x - 3)(2 - 3x)(5 - x)# ?

1 Answer
Oct 5, 2014

The product rule states that #(hfg)'=h'fg+hf'g+hfg'#.

So, #(d/dx(5-x))(x-3)(2-3x)+(5-x)(d/dx(x-3))(2-3x)+(5-x)(x-3)(d/dx(2-3x))#

The derivative of #5-x# is #-1# since the constant has a derivative of o and the derivative of #-x# is #-1# from #(1)*-1x^(1-1)# giving #-1x^0 or -1.#

The derivative of #x-3# is 1 as above.

The derivative of #2-3x# is -3 from #1*-3x^(1-1)=-3^0#

Substituting back we get #=(-1)((x-3))((2-3x))+((5-x))(1)((2-3x))+((5-x))((x-3))(-3)#

Using FOIL we get #3x^2-11x+6+3x^2-17x+10+3x^2-24x+45.#

Collecting like terms we get our derivative; #9x^2-52x+61.#