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

If `y=(ax^2)/((x-a)(x-b)(x-c))+(b x)/((x-b)(x-c))+c/(x-c)+1`, then prove that `(y')/y=1/x[a/(a-x)+b/(b-x)+c/(c-x)]`

Text Solution

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`y=(ax^(2))/((x-a)(x-b)(x-c))+(bx)/((x-b)(x-c))+(c)/(x-c)+1`
`=(ax^(2))/((x-a)(x-b)(x-c))+(bx)/((x-b)(x-c))+((c+x-c)/(x-c))`
`=(ax^(2))/((x-a)(x-b)(x-c))+(bx)/((x-b)(x-c))+(x)/(x-c)`
`=(ax^(2))/((x-a)(x-b)(x-c))+(bx+x(x-b))/((x-b)(x-c))`
`=(ax^(2))/((x-a)(x-b)(x-c))+(x^(2))/((x-b)(x-c))`
`=(ax^(2)+x^(2)(x-a))/((x-a)(x-b)(x-c))`
`=(x^(3))/((x-a)(x-b)(x-c))`
`therefore" "log y= log {(x^(3))/((x-a)(x-b)(x-c))}`
`"or "log y=3log x-{log(x-a)+log(x-b)+log(x-c)}`
On differentiating w.r.t. x, we get
`(1)/(y)(dy)/(dx)=(3)/(x)-{(1)/(x-a)+(1)/(x-b)+(1)/(x-c)}`
`"or "(dy)/(dx)=y{((1)/(x)-(1)/(x-a))+((1)/(x)-(1)/(x-b))+((1)/(x)-(1)/(x-c))}`
`=y{-(a)/(x(x-a))-(b)/(x(x-b))-(c)/(x(x-c))}`
`=(y)/(2){(a)/(a-x)+(b)/(b-x)+(c)/(x-c)}.`
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