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Question Number 105613 by Study last updated on 30/Jul/20
f(x)=(1+(1/x))^(x!) f^′ (x)=????
f(x)=(1+1x)x!f′(x)=????
Answered by mathmax by abdo last updated on 30/Jul/20
f(x) =e^(x!ln(1+(1/x))) ⇒f^′ (x) =(x!ln(1+(1/x)))^′ f(x) le =(x!)^′ ln(1+(1/x))+x!(ln(1+(1/x)))^′ f(x) but x!=Γ(x+1) =∫_0 ^∞ t^x e^(−t) dt =∫_0 ^∞ e^(xlnt) e^(−t) dt ⇒ (d/dx)(x!) = ∫_0 ^∞ lnt t^x e^(−t) dt and (d/dx)(ln(1+(1/x)))=((−1)/(x^2 (1+(1/x)))) =((−1)/(x^2 +x)) ⇒f^′ (x) ={ln(1+(1/x))∫_0 ^∞ t^x e^(−t) lnt dt −((x!)/(x^2 +x))}(1+(1/x))^(x!)
f(x)=ex!ln(1+1x)⇒f′(x)=(x!ln(1+1x))′f(x)le=(x!)′ln(1+1x)+x!(ln(1+1x))′f(x)butx!=Γ(x+1)=∫0∞txe−tdt=∫0∞exlnte−tdt⇒ddx(x!)=∫0∞lnttxe−tdtandddx(ln(1+1x))=−1x2(1+1x)=−1x2+x⇒f′(x)={ln(1+1x)∫0∞txe−tlntdt−x!x2+x}(1+1x)x!

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