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Question Number 2231 by Filup last updated on 10/Nov/15
For f(x)=x!  Is there a defined:    (1) f′(x)  (2) ∫f(x)dx
$$\mathrm{For}\:{f}\left({x}\right)={x}! \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{defined}: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{f}'\left({x}\right) \\ $$$$\left(\mathrm{2}\right)\:\int{f}\left({x}\right){dx} \\ $$
Commented by Filup last updated on 10/Nov/15
Is this a correct assumption for (1):    y=x!  y=x(x−1)(x−2)...  ∴y=Π_(i=1) ^x i    Using Chain rule:  ∴y′=[(x−1)(x−2)...]+[x(x−2)...]+...  y′=(y/x)+(y/(x−1))+...=y((1/x)+(1/(x−1))+...+1)  y′=yΣ_(i=1) ^(x−1) ((1/i))  ∴y′=y∙H(x−1)             where H(n) is the sum to the nth              harmonic term  y=Π_(i=1) ^x i  y′=yΣ_(i=1) ^(x−1) (1/i)    Note: I have no knowledge of differentiating  this function, so this is only a guess attempt
$$\mathrm{Is}\:\mathrm{this}\:\mathrm{a}\:\mathrm{correct}\:\mathrm{assumption}\:\mathrm{for}\:\left(\mathrm{1}\right): \\ $$$$ \\ $$$${y}={x}! \\ $$$${y}={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)… \\ $$$$\therefore{y}=\underset{{i}=\mathrm{1}} {\overset{{x}} {\prod}}{i} \\ $$$$ \\ $$$${Using}\:{Chain}\:{rule}: \\ $$$$\therefore{y}'=\left[\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…\right]+\left[{x}\left({x}−\mathrm{2}\right)…\right]+… \\ $$$${y}'=\frac{{y}}{{x}}+\frac{{y}}{{x}−\mathrm{1}}+…={y}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}}+…+\mathrm{1}\right) \\ $$$${y}'={y}\underset{{i}=\mathrm{1}} {\overset{{x}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}}{{i}}\right) \\ $$$$\therefore{y}'={y}\centerdot\mathrm{H}\left({x}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{where}\:\mathrm{H}\left({n}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{the}\:{n}\mathrm{th} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{harmonic}\:\mathrm{term} \\ $$$${y}=\underset{{i}=\mathrm{1}} {\overset{{x}} {\prod}}{i} \\ $$$${y}'={y}\underset{{i}=\mathrm{1}} {\overset{{x}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{i}} \\ $$$$ \\ $$$$\mathrm{Note}:\:\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{knowledge}\:\mathrm{of}\:\mathrm{differentiating} \\ $$$$\mathrm{this}\:\mathrm{function},\:\mathrm{so}\:\mathrm{this}\:\mathrm{is}\:\mathrm{only}\:\mathrm{a}\:\mathrm{guess}\:\mathrm{attempt} \\ $$
Commented by 123456 last updated on 10/Nov/15
f(x)=x!=Γ(x+1)  f′(x)=ψ(x+1)x!  ψ(x)=(d/dx)ln Γ(x)=((Γ′(x))/(Γ(x)))
$${f}\left({x}\right)={x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$${f}'\left({x}\right)=\psi\left({x}+\mathrm{1}\right){x}! \\ $$$$\psi\left({x}\right)=\frac{{d}}{{dx}}\mathrm{ln}\:\Gamma\left({x}\right)=\frac{\Gamma'\left({x}\right)}{\Gamma\left({x}\right)} \\ $$

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