Question

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

Answer 1

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.`