How do differentiate a function f(x)=x!?

Question

if a function f(x)=x!, a factorial of x,
how to get f'(x)?

Note: n! = 1x2x3x4x.....xn

by the way, I believe it must has some solution, I need it badly.
f'(x) is the slope of f(x)=x!. You can imagine that, the value of f'(x) going to very big follows by x (x is no less than 0, anyway). But it won't bigger than f'(y) where f(y)=y ^ y.
thus, I belive f'(x) < x^x ln(x),
but don't know what exactly it is.

Answer 1

x! is only defined for integer x, so it has no derivative. However, there is a related function, called the Gamma function Gamma(x) which is differentiable. Unfortunately, the derivative of Gamma is rather nasty. If, as I suspect, you want the derivative so you can apply L'hopital's rule to find convergence of a series by a limit test, I would suggest that you use the ratio test instead.

Answer 2

We can rewrite factorials in series form as
f(x)=X!
= (x-0)(x-1)(x-2)...(x-(n-1))
where n is the number of terms. Now, let us take only the first 3 terms and try to differentiate.
f(x) (for first 3 terms) = (x-0)(x-1)(x-2) = x^3 - 3x^2 + 2x
f(3) = 27-27+6 = 6 = 3*2*1 = 3!
f'(x) = 3x^2 - 6x + 2
If we take the first 5 terms,
f'(x) (for 5 terms) = 5x^4-40x^3+105x^2-100x^2+24
In this manner if we can go on we can finally differentiate the polynomial with degree n. There is a series equivalent of X!which I do not remember but similar to the above.

Answer 3

Factorials can be defined for non-integers. From wikipedia:

"The factorial function can also be defined for non-integer values, but this requires more advanced tools from mathematical analysis. The function that "fills in" the values of the factorial between the integers is called the Gamma function..." The gamma function has continuous segments which in theory can be differentiated.

See http://en.wikipedia.org/wiki/Factorials

Answer 4

i will assume those are the partials with comprehend to x, y, z and t. enable's do fx. So manage y, z and t as constants. meaning the denominator (t + 7z) is a continuing and so is y^8. so that you've x * a/b the position a and b are constants. The by-product is only a/b = y^8/(t + 7z). For fy, imagine of it as y^8 * a/b For fz, you've were given the shape a/(b + 7z) For ft, you've were given a/(t + b) In each case, i'm lumping all the different variables mutually as some consistent a or b. The words a and b stand for diverse issues in each case.

Answer 5

The Gamma function, \Gamma(x) satisfies \Gamma(n)=n!

The Gamma function is defined by

\Gamma(x)=\int_{0}^{\infty} (t^(x-1) e^(-t)) dt

For real values of x, it is differentiable for x >0. It is discontinuous at the _negative_ integers.

The derivative of the Gamma function is

\Gamma'(x)=-\Gamma(x) (1/x+\gamma+
\sum_{n=1}^{\infty}(1/(n+x)-
1/n)),
where \gamma is the Euler-Mascheroni constant. This contant arises in numerous integrals and is approximately 0.5772156649. It is defined by the improper integral
\gamma=-\int_{0}^{\infty} (e^(-x) ln(x) ) dx.

Answer 6

x can be a negative or positive value, therefore depend on what ur x value is first, then u can take the factorial by the power of multiplication of 2.

Answer 7

I think the x is a number of cource which is a constant.
And hence the answer may b 0.

Answer 8

well....i suppose that a factorial IS a number as in, itd a constant. so this means that the derivative is going to be 0.

Answer 9

Here are some useful links regarding the generalized factorial function (Gamma functon) and its derivative.

http://www.sosmath.com/calculus/improper/gamma/gamma.html

http://mathforum.org/library/drmath/view/53664.html

http://en.wikipedia.org/wiki/Factorial

Answer 10

f'(x) is defined as limit h tends to zero for (f(x+h)-f(x))/h,
however for f(x) = x!,
1.function is not defined at intermediate values, but whole numbers. function may be defined at whole number x, but not x+h
2.Function is not continous, hence the derivate doesnt exists