Question

Differentiate `-x^2-2x-2` using first principles?

Answer 1

`d/dx( -x^2-2x-2) = -2x-2`

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

Which we sometimes refer to as the "difference quotient" . So with `f(x) = -x^2-2x-2` then;

`f(x+h) = -(x+h)^2-2(x+h)-2`
`" " =-(x^2+2xh+h^2) -2x-2h-2`
`" " =-x^2-2xh-h^2 -2x-2h-2`

And so:

`f(x+h) - f(x) = -x^2-2xh-h^2 -2x-2h-2 - (-x^2-2x-2 )`
`" " = -x^2-2xh-h^2 - 2x - 2h - 2 + x^2 +2x+2`
`" " = -x^2-2xh-h^2 -2h + x^2`

And so the derivative of `y=f(x)` is given by:

`f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
`\ \ \ \ \ \ \ \ \= lim_(h rarr 0) ( -x^2-2xh-h^2 -2h + x^2 ) / h`
`\ \ \ \ \ \ \ \ \= lim_(h rarr 0) ( -x^2/h - 2x - h -2 + x^2 / h )`
`\ \ \ \ \ \ \ \ \= -0 - 2x - 0 -2 + 0`
`\ \ \ \ \ \ \ \ \= - 2x -2`