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"If "=(ax^(2))/((x-a)(x-b)(x-c))+(bx)/((...

`"If "=(ax^(2))/((x-a)(x-b)(x-c))+(bx)/((x-b)(x-c))+(c)/(x-c)+1," prove that "(y')/(y)=(1)/(x)((a)/(a-x)+(b)/(b-x)+(c)/(c-x)).`

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`y=(ax^(2))/((x-a)(x-b)(x-c))+(bx)/((x-b)(x-c))+c/((x-c))+1`
`y=(ax^(2))/((x-a)(x-b)(x-c))+(bx)/((x-b)(x-c))+x/((x-c))`
`=(ax^(2))/((x-a)(x-b)(x-c))+x/((x-c))(b/(x-b)+1)`
`=(ax^(2))/((x-a)(x-b)(x-c))+x/((x-c))*x/((x-b))`
`=x^(2)/((x-c)(x-b))(a/(x-1)+1)rArr y= x^(3)/((x-a)(x-b)(x-c))`
` rArr log y = log x^(3) - log (x-a) (x-b)(x-c)`
` rArr log y = 3 log x - log (x-a) - log (x-b) - log (x-c)`
On differentiating, we get
`(y')/y = 3/x - 1/(x-a) -1/(x-b)-1/(x-c) `
`rArr" "(y')/y =(1/x-1/(x-a))+(1/x-1/(x-b))+(1/x-1/(x-c))`
` rArr" "(y')/y = (-a)/(x(x-a))-b/(x(x-b))-c/(x(x-c)) `
`rArr" " (y')/y = a/(x(a-x))+b/(x(b-x))+c/(x(c-x))`
` rArr" "(y')/y = 1/x(a/(a-x)+b/(b-x)+c/(c-x))`
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