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Question Number 165641 by nadovic last updated on 05/Feb/22

$$\mathrm{Given}\mathrm{that}y=\frac{1}{x}$$$$\left(a\right)\mathrm{Show}\mathrm{that}{y}^{\left(n\right)}=\frac{{(-1)}^{n}n!}{x^{n+1}}$$$$\left(b\right)\mathrm{Find}\mathrm{an}\mathrm{expression}\mathrm{for}{y}^{(n-1)}+y^{\left(n\right)}$$$$$$

Answered by aleks041103 last updated on 05/Feb/22

$$a)$$$$y^{\left(0\right)}=\frac{{(-1)}^{0}0!}{x^{0+1}}=\frac{1}{x}=y✓$$$$suppose$$$$y^{\left(k\right)}=\frac{{(-1)}^{k}k!}{x^{k+1}}$$$$then$$$$\text{Prime causes double exponent: use braces to clarify}$$$$={(-1)}^{k}k!{\left(x^{-k-1}\right)}^{\prime }=$$$$={(-1)}^{k}k!(-k-1)x^{-k-2}=$$$$={(-1)}^{k+1}k!(k+1)\frac{1}{x^{k+2}}=$$$$=\frac{{(-1)}^{k+1}(k+1)!}{x^{(k+1)+1}}✓$$$$\Rightarrow byinduction:$$$$\mathrm{\forall }n\in {\mathbb{N}}^0,y^{\left(n\right)}=\frac{d^n}{{dx}^n}\left(\frac{1}{x}\right)=\frac{{(-1)}^{n}n!}{x^{n+1}}$$

Answered by aleks041103 last updated on 05/Feb/22

$$b)$$$$y^{(n-1)}=\frac{{(-1)}^{n-1}(n-1)!}{x^n}$$$$y^{\left(n\right)}=\frac{{(-1)}^{n}n!}{x^{n+1}}$$$$\Rightarrow y^{(n-1)}+y^{\left(n\right)}=\frac{{(-1)}^{n}n!}{x^{n+1}}(1+\frac{(-1)x}{n})$$$$y^{(n-1)}+y^{\left(n\right)}=\frac{{(-1)}^n(n-1)!}{x^{n+1}}(n-x)$$

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