Is there any short way for derivative of this function: f(x)=(x−1)(x−2)(x−3)(x−4)(x−5)?

Answer 1

This can be solved by using both expanding and the product rule:

I will first take the first three terms (x-1)(x-2)(x-3) and expand and I would call their product u

I will take the the last two terms (x-4)(x-5) and expand them and I would call their product v

using the product rule for one time only the derivative would be u'v+v'u . eventually you will have to expand the result which will be a polynomial also of probably the fifth degree. I will leave the math for you to do.

Answer 2

The best way to solve a problem like this is not to attempt to recursively use the product rule, but rather to expand the expression into a polynomial. Once you have the right-hand side in the form of a polynomial, you can take the derivative of each term as a simple expression like ax^n. Polynomial expansion is tedious, but easier than keeping track of the product rule over multiple products, because even if you use the product rule you'll end up expanding it into a polynomial to find the simplified solution, but you'll do a polynomial expansion for each product term (which is a lot more work).

Answer 3

Multiply them all together:

(x^2 - 3x + 2)(x^2 - 7x + 12)(x-5)

[x^4 - 7x^3 + 12x^2

-3x^3 + 21x^2 - 36x

2x^2 - 14x + 24] (x+5)

[x^4 - 10x^3 + 35x^2 - 50x + 24](x+5)

x^5 - 10x^4 + 35x^3 - 50x^2 + 24x

5x^4 - 50x^3 + 175x^2 - 250x + 120

x^5 - 5x^4 - 15x^3 + 125x^2 - 226x + 120

The derivative is 5x^4 - 20x^3 - 45x^2 + 250x - 226

Answer 4

Let a=x-1; b=x-2; c=x-3; d=x-4; e=x-5

d(f(x))/dx = d(abcde)/dx = abcd*de/dx + abce*dd/dx + abde*dc/dx + acde*db/dx + bcde*da/dx

Since da/dx = db/dx = dc/dx = dd/dx = de/dx = 1

d(f(x))/dx = abcd + abce + abde + acde + bcde

= (x-1)(x-2)(x-3)(x-4) + (x-1)(x-2)(x-3)(x-5) + (x-1)(x-2)(x-4)(x-5) +

(x-1)(x-3)(x-4)(x-5) + (x-2)(x-3)(x-4)(x-5)

Answer 5

In order to use the product rule, you will have to FOIL and multiply pairs of them anyway.

Go ahead and multiply them all together.

You can differentiate the 5th degree polynomial using the power rule

Answer 6

If there is, I sure don't know it.

I think probably the simplest and surest way is to multiply the binomials together to get a 5th degree polynomial which is then easy to differentiate.

Answer 7

Let's look at the product rule of two functions for a moment.

(fg)' = f' g + f g'

Note that we have fg twice (well almost!) since there are two functions. Note that in the 1st term we take the derivative of f and in the 2nd term we take the derivative of g.

It can be shown by induction that, for example, (fghij)' = (f'ghij) + (fg'hij) + (fgh'ij) + (fghi'j) + (fghij')

Answer 8

The easiest way would be to use the product rule. Let:

s = (x – 1)

t = (x – 2)

u = (x – 3)

v = (x – 4)

w = (x – 5)

Then:

f' = s'tuvw + st'uvw + stu'vw + stuv'w + stuvw'

= (x – 2)(x – 3)(x – 4)(x – 5) + (x – 1)(x – 3)(x – 4)(x – 5) + (x – 1)(x – 2)(x – 4)(x – 5) + (x – 1)(x – 2)(x – 3)(x – 5) + (x – 1)(x – 2)(x – 3)(x – 4)

If your teacher wants you to expand all of that, they're not very nice.

Answer 9

It's really not hard to multiply out. I did it in just a couple of minutes and got:

f(x) = x⁵ - 15x⁴ + 85x³ - 225x² + 274x - 120

Then the derivative (by the power rule) is:

f'(x) = 5x⁴ - 60x³ + 255x² - 450x + 274