Question

How do you find the first and second derivative of `x^2-2x-3`?

Answer 1

If `f(x)=x^2-2x-3`
`color(white)("XXX")f'(x)=2x-2` and
`color(white)("XXX")f''(x)=2`

Explanation:

The first derivative:
`f'(x)=(df(x))/(dx)=(dcolor(red)(x^2))/(dx)-(dcolor(blue)(2x))/(dx)-(dcolor(green)(3))/(dx)`

`color(white)("XXXXXXXXX")=color(red)(2x^1)-color(blue)(2x^0)-color(green)(0)`

`color(white)("XXXXXXXXX")=color(red)(2x)-color(blue)(2)`

The second derivative
The second derivative is just the derivative of the first derivative.
`f''(x)=(df'(x))/(dx) = (dcolor(red)(2x^1))/(dx)-(dcolor(blue)(2x^0))/(dx)`

`color(white)("XXXXXXXXXX")=color(red)(2x^0)-color(blue)(0)`

`color(white)("XXXXXXXXXX")=color(red)(2)`