Question

Last derivatives based problem. Help would be appreciated.

Answer 1

The derivative of
`f(x)` is defined by the limit,

`f'(x)=\displaystyle\lim_(h\to0)\frac{f(x+h)-f(x)}h`

In this case,
`f(x)=8x-x^2-4`, so the derivative would be

`f'(x)=\displaystyle\lim_(h\to0)\frac{(8(x+h)-(x+h)^2-4)-(8x-x^2-4)}h`

Simplifying the numerator gives

`\frac{(8x+8h-x^2-2xh-h^2-4)-(8x-x^2-4)}h=\frac{8h-2xh-h^2}h`

The numerator and denominator share a common factor of
`h`, which we can cancel because
`h\to0` means that we're considering
`h\\eq0` so that
`\frac hh=1`:

`\frac{8h-2xh-h^2}h=8-2x-h`

Then as
`h\to0`, we're left with the derivative

`f'(x)=8-2x`