Calculate-d-n-dx-n-sinx-d-n-dx-n-lnx-

Question Number 108350 by Ar Brandon last updated on 16/Aug/20

$$\mathrm{Calculate}\:\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{dx}^{\mathrm{n}} }\left(\mathrm{sinx}\right)\:,\:\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{dx}^{\mathrm{n}} }\left(\mathrm{lnx}\right) \\ $$

Answered by Dwaipayan Shikari last updated on 16/Aug/20

f(x)=logx f′(x)=(1/x) f′′(x)=−(1/x^2 ) f′′′(x)=(2/x^3 ) f^((iv)) =−(6/x^4 ) f^((v)) =((24)/x^5 ) ...f^((n)) =(−1)^(n+1) .(((n−1)!)/x^n )

$${f}\left({x}\right)={logx} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$$${f}''\left({x}\right)=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${f}'''\left({x}\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} } \\ $$$${f}^{\left({iv}\right)} =−\frac{\mathrm{6}}{{x}^{\mathrm{4}} } \\ $$$${f}^{\left({v}\right)} =\frac{\mathrm{24}}{{x}^{\mathrm{5}} } \\ $$$$…{f}^{\left({n}\right)} =\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} .\frac{\left({n}−\mathrm{1}\right)!}{{x}^{{n}} } \\ $$

Answered by ajfour last updated on 16/Aug/20

y=sin x (dy/dx)=y_1 =sin ((π/2)+x) y_2 =sin [2((π/2))+x] .... y_n =sin [n((π/2))+x]

$${y}=\mathrm{sin}\:{x} \\ $$$$\frac{{dy}}{{dx}}={y}_{\mathrm{1}} =\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{\mathrm{2}} =\mathrm{sin}\:\left[\mathrm{2}\left(\frac{\pi}{\mathrm{2}}\right)+{x}\right] \\ $$$$…. \\ $$$${y}_{{n}} =\mathrm{sin}\:\left[{n}\left(\frac{\pi}{\mathrm{2}}\right)+{x}\right] \\ $$

Answered by mathmax by abdo last updated on 17/Aug/20

(d^n /dx^n )(sinx) =sin(x+((nπ)/2)) we can prove this by recurrence (d/dx)(lnx) =(1/x) ⇒(d^n /dx^n )(lnx) =((1/x))^((n−1)) =(((−1)^(n−1) (n−1)!)/x^n ) for n≥1

$$\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{dx}^{\mathrm{n}} }\left(\mathrm{sinx}\right)\:=\mathrm{sin}\left(\mathrm{x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right)\:\mathrm{we}\:\mathrm{can}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{by}\:\mathrm{recurrence} \\ $$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{lnx}\right)\:=\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{dx}^{\mathrm{n}} }\left(\mathrm{lnx}\right)\:=\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\left(\mathrm{n}−\mathrm{1}\right)} \:=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{x}^{\mathrm{n}} } \\ $$$$\mathrm{for}\:\mathrm{n}\geqslant\mathrm{1} \\ $$

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