how do I find the derivative of this??

Question

find f'(x) for the function below with the product rule, rather than multiplying out.

f(x) = (x-1)(x-2)(x-3)


f(x) = (x-1)(x-2)(x-3)(x-4)

the question asks me not to use the product rule directly!!!!! I want to but I cant

Answer 1

Well, you could use the multiplication rule but I would suggest to just multiply it out first:

f(x) = (x-1)(x-2)(x-3) = (x²-3x+1)(x-3) = (x³-3x²-3x³+9x+x-3)
f(x) = -2x³-3x²+10x-3

Then find the derivative:
f'(x) = -6x²-6x+10

The second problem do the same
f(x) = (x-1)(x-2)(x-3)(x-4) = (x³-3x²-3x³+9x+x-3)(x-4)
.... etc

I think you can take it from here

Answer 2

If you are at the calculus level, which you are, then I can avoid doing all the algebra for you, and just tell you that:

Multiply everything out in the parentheses, and you will have a 3rd order polynomial for the first one, anda 4th order poly for the second one. Once you have that you can take the derivative by sight.

Answer 3

f(x) = (x-1)(x-2)(x-3)(x-4)
f'(x)={d(x-1)/dx}(x-2)(x-3)(x-4)+{d(x-2)/dx}(x-1)(x-3)(x-4)+{d(x-3)/dx}(x-1)(x-2)(x-4)+{d(x-4)/dx}(x-1)(x-2)(x-3)

f'(x)=(x-2)(x-3)(x-4)+(x-1)(x-3)(x-4)+
(x-1)(x-2)(x-4)+(x-1)(x-2)(x-3)