Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 108043 by ZiYangLee last updated on 14/Aug/20

Given f(x)=x(x+1)(x+2)...(x+n) find the value of f′(0).

$$\mathrm{Given}\:\mathrm{f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)...\left({x}+{n}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}'\left(\mathrm{0}\right). \\ $$

Answered by hgrocks last updated on 14/Aug/20

ln(f(x)) = Σ_(k=0) ^n ln(x+k) ((f′(x))/(f(x))) = (1/x) + (1/(x+1)) +(1/(x+2)) +.......(1/(x+n)) ((f′(0))/(f(0))) = (1/x) + 1 + (1/2) +(1/3)+.........(1/n) f′(0) = Π_(k=1) ^n (x+k)_(x=0) + f(0)(H_n ) = 1.2.3.....n + 0 × H_n = n!

$$\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{x}+\mathrm{k}\right) \\ $$$$\frac{\mathrm{f}'\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2}}\:+.......\frac{\mathrm{1}}{\mathrm{x}+\mathrm{n}} \\ $$$$\frac{\mathrm{f}'\left(\mathrm{0}\right)}{\mathrm{f}\left(\mathrm{0}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}+.........\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{f}'\left(\mathrm{0}\right)\:=\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{x}+\mathrm{k}\right)_{\mathrm{x}=\mathrm{0}} \:+\:\mathrm{f}\left(\mathrm{0}\right)\left(\mathrm{H}_{\mathrm{n}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}.\mathrm{2}.\mathrm{3}.....\mathrm{n}\:+\:\mathrm{0}\:×\:\mathrm{H}_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{n}! \\ $$$$ \\ $$

Commented by ZiYangLee last updated on 14/Aug/20

wow...

$$\mathrm{wow}... \\ $$

Answered by Her_Majesty last updated on 14/Aug/20

f(x)=x(x+1)(x+2)...(x+n)= =x^(n+1) +(....)x^2 +n!x f′(x)=(n+1)x^n +2(...)x+n! f′(0)=n!

$${f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)...\left({x}+{n}\right)= \\ $$$$={x}^{{n}+\mathrm{1}} +\left(....\right){x}^{\mathrm{2}} +{n}!{x} \\ $$$${f}'\left({x}\right)=\left({n}+\mathrm{1}\right){x}^{{n}} +\mathrm{2}\left(...\right){x}+{n}! \\ $$$${f}'\left(\mathrm{0}\right)={n}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com