Question #3fc87

2 Answers
Apr 11, 2016

#F'(a) = (a-b)(a-c)(a-d)(a-e)(a-f)#


#F' = (x-a) [(x-b)(x-c)(x-d)(x-e)(x-f)]' + (x-b)(x-c)(x-d)(x-e)(x-f)#
Evaluated at #x=a# the first term vanishes. The second is the answer

Apr 12, 2016

Let me try to explain more


Write F as (x-a) times G(x) where G is all the factors of F except the first one.

#F(x) = (x-a) G(x)#

its derivative is (using the product rule)

F'(x) = (x-a) G'(x) + (x-a)'G(x)

so far so good.

Now for the value x = a, the first term vanishes, The derivative of #(x-a)# is #1# and the last term of the total derivative is G(a) but G(a) is nothing but the original F divided by x-a and with all x replaced with a.

F(x) = (x-a)(x-b)(x-c)(x-d)(x-e)(x-f)
F'(a) = (a-b)(a-c)(a-d)(a-e)(a-f)

That is a very simple expression to remember, but not so easy to derive!