# Question #3fc87

Apr 11, 2016

$F ' \left(a\right) = \left(a - b\right) \left(a - c\right) \left(a - d\right) \left(a - e\right) \left(a - f\right)$

#### Explanation:

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

Apr 12, 2016

Let me try to explain more

#### Explanation:

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

$F \left(x\right) = \left(x - a\right) G \left(x\right)$

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 $\left(x - a\right)$ 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!