Question

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

Answer 1

`h^' = (s-2)/s^3`

Explanation:

You can differentiate this function by using the quotient rule, which allows you to differentiate functions that take the form

`color(blue)(h(x) = f(x)/g(x))" "`, where `color(blue)(g(x)!=0)`

by using the formula

`color(blue)(d/dx(h(x)) = ([d/dx(f(x))] * g(x) - f(x) * d/dx(g(x)))/(g(x))^2)`

In your case, you have

`f(s) = 3s^2 - 2 + 1" "` and `" "g(s) = s^2`

This means that you can write

`d/(ds)(h(s)) = ([d/(ds)(3s^2 - s + 1) * s^2 - (3s^2 - s + 1) * d/dx(s^2)))/(s^2)^2`

`h^' = ((6s - 1) * s^2 - (3s^2 - s + 1) * 2s)/s^4`

`h^' = (color(red)(cancel(color(black)(6s^3))) - s^2 - color(red)(cancel(color(black)(6s^3))) + 2s^2 - 2s)/s^4`

`h^' = (s^2 - 2s)/s^4`

You can simplify this further to get

`h^' = (color(red)(cancel(color(black)(s))) * (s - 2))/s^color(red)(cancel(color(black)(4))) = color(green)((s-2)/s^3)`