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Question Number 33407 by NECx last updated on 15/Apr/18
if y=x! find dy/dx
$${if}\:{y}={x}!\:{find}\:{dy}/{dx} \\ $$
Answered by MJS last updated on 15/Apr/18
for x∈N no derivate exists  if we use  y=x!=Γ(x)=∫_0 ^∞ (t^x /e^t )dt (x∈R) ⇒  ⇒ (dy/dx)=Γ′(x)=∫_0 ^∞ ((t^x ln t)/e^t )dt  if we use  y=x!=Γ(x)=∫_0 ^1 (ln (1/t))^x dt (x∈R) ⇒  ⇒ (dy/dx)=Γ′(x)=∫_0 ^1 (ln (1/t))^x ln (ln (1/t))dt
$$\mathrm{for}\:{x}\in\mathbb{N}\:\mathrm{no}\:\mathrm{derivate}\:\mathrm{exists} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{use} \\ $$$${y}={x}!=\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}^{{x}} }{\mathrm{e}^{{t}} }{dt}\:\left({x}\in\mathbb{R}\right)\:\Rightarrow \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}=\Gamma'\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}^{{x}} \mathrm{ln}\:{t}}{\mathrm{e}^{{t}} }{dt} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{use} \\ $$$${y}={x}!=\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{ln}\:\frac{\mathrm{1}}{{t}}\right)^{{x}} {dt}\:\left({x}\in\mathbb{R}\right)\:\Rightarrow \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}=\Gamma'\left({x}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{ln}\:\frac{\mathrm{1}}{{t}}\right)^{{x}} \mathrm{ln}\:\left(\mathrm{ln}\:\frac{\mathrm{1}}{{t}}\right){dt} \\ $$
Commented by prof Abdo imad last updated on 15/Apr/18
we know that Γ(x) =∫_0 ^∞  t^(x−1) e^(−t) dt with x>0  and Γ(x+1) =xΓ(x) so if we note  x!=Γ(x+1) we get x! =∫_0 ^∞   t^x  e^(−t) dt  = ∫_0 ^∞  e^(xln(t))  e^(−t) dt ⇒ (d/dx)(x!) = ∫_0 ^∞ (∂/∂x)( e^(xln(t))  e^(−t) )dt  = ∫_0 ^∞   e^(−t) ln(t) t^x  dt
$${we}\:{know}\:{that}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}\:{with}\:{x}>\mathrm{0} \\ $$$${and}\:\Gamma\left({x}+\mathrm{1}\right)\:={x}\Gamma\left({x}\right)\:{so}\:{if}\:{we}\:{note} \\ $$$${x}!=\Gamma\left({x}+\mathrm{1}\right)\:{we}\:{get}\:{x}!\:=\int_{\mathrm{0}} ^{\infty} \:\:{t}^{{x}} \:{e}^{−{t}} {dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{{xln}\left({t}\right)} \:{e}^{−{t}} {dt}\:\Rightarrow\:\frac{{d}}{{dx}}\left({x}!\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\partial}{\partial{x}}\left(\:{e}^{{xln}\left({t}\right)} \:{e}^{−{t}} \right){dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left({t}\right)\:{t}^{{x}} \:{dt}\: \\ $$

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