Đạo hàm cấp cao

R

rubitaku12

N

nhocngo976

Giúp mình bài này với:)
Tính đạo hàm cấp n của hàm số: [tex]y=\frac{5x-3}{x^2-3x+2}[/tex]


[TEX]y=\frac{-2}{(x-1)}+\frac{7}{(x-2)}[/TEX]

[TEX]y'=\frac{2}{(x-1)^2 } -\frac{7}{(x-2)^2[/TEX]

[TEX]y''= \frac{-2.2}{(x-1)^3}+\frac{7.2}{(x-2)^3}=-1.2.[\frac{2}{(x-1)^3}-\frac{7}{(x-2)^2}][/TEX]

[TEX]y_{3}=(-1)^2.2.3[\frac{2}{(x-1)^4 -\frac{7}{(x-2)^4}][/TEX]

\Rightarrow[TEX]y^{(n)}=(-1)^{n-1}.n!.[\frac{2}{(x-1)^{n-1}}-\frac{7}{(x-2)^{n-1}}](*)[/TEX]

[TEX]chung \ minh \ (*) \ = \ pp \ quy \ nap [/TEX][TEX][/TEX]
 
N

nhocngo976

[tex]lim\frac{1-cosxcos3x}{x^2}[/tex] x->0


[TEX]\lim_{x\to0}\frac{1-(1-sin^2x)(1-sin^23x)}{(1+cosxcos3x)x^2}=\lim_{x\to0}\frac{sin^23x+sin^2x-sin^2xsin^23x}{(1+cosxcos3x)x^2}=\lim_{x\to0}\frac{\frac{sin^23x.9}{(3x)^2}+\frac{sin^2x}{x^2}-\frac{sin^2x}{x^2}sin^23x}{1+cosxcos3x}=\frac{9+1-0}{2}=5[/TEX][TEX][/TEX][TEX][/TEX][TEX][/TEX][TEX][/TEX][TEX][/TEX]
 
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