Question

How do you find the derivative of `x=((s-1)(s-2))/(s-3)`?

Answer 1

`(dx)/(ds)=(s^2-6s+7)/(s-3)^2`

Explanation:

According to Quotient rule if `f(x)=(g(x))/(h(x))`

then `(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2`

Here we have `x=((s-1)(s-2))/(s-3)=(s^2-3s+2)/(s-3)`

Hence `(dx)/(ds)=((2s-3)(s-3)-(s^2-3s+2)xx1)/(s-3)^2`

= `(2s^2-9s+9-s^2+3s-2)/(s-3)^2`

= `(s^2-6s+7)/(s-3)^2`