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

$\dfrac{(8x+8h-x^2-2xh-h^2-4)-(8x-x^2-4)}h=\dfrac{8h-2xh-h^2}h$

The numerator and denominator share a common factor of $h$, which we can cancel because $h\to0$ means that we're considering $h\neq0$ so that $\dfrac hh=1$:

$\dfrac{8h-2xh-h^2}h=8-2x-h$

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

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