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Differentiate the function with respect ...

Differentiate the function with respect of `x \ :1/(a \ x^2+b \ x+c)`

Text Solution

Verified by Experts

Let u=1 and` v=ax^2+bx+c`
Then (du/dx)=0 and (dv/dx)=2ax+b
As by quotient rule
`(dy/dx)=(v(du/dx)-u(dv/dx))/v^2`
So putting the values we get
`(dy/dx)=-(2ax+b)/(ax^2+bx+c)^2`
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