Using the limit definition, how do you find the derivative of $f\left( x \right) = 4 - 2x - {x^2}$?

Hint: We need to plug this function into the definition of the derivative and do some algebra. First plug the function into the definition of the derivative. Next, multiply everything out and distribute the minus sign through on the second term. Notice that every term in the numerator that didn’t have an $h$ in it cancelled out and we can now factor an $h$ out of the numerator which will cancel against the $h$ in the denominator. After that we can compute the limit.

Formula used:
Definition of the Derivative
The derivative of $f\left( x \right)$ with respect to $x$ is the function $f'\left( x \right)$ and is defined as,
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$…(1)

Complete step by step solution:
So, all we really need to do is to plug this function into the definition of the derivative, (1), and do some algebra.
First plug the function into the definition of the derivative.
 $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{4 - 2\left( {x + h} \right) - {{\left( {x + h} \right)}^2} - \left( {4 - 2x - {x^2}} \right)}}{h}$
Be careful and make sure that you properly deal with parenthesis when doing the subtracting.
Now, we know that we can’t just plug in $h = 0$ since this will give us a division by zero error. So, we are going to have to do some work. In this case that means multiplying everything out and distributing the minus sign through on the second term. Doing this gives,
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{4 - 2x - 2h - {x^2} - {h^2} - 2xh - 4 + 2x + {x^2}}}{h}$
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2h - {h^2} - 2xh}}{h}$
Notice that every term in the numerator that didn’t have an $h$ in it cancelled out and we can now factor an $h$ out of the numerator which will cancel against the $h$ in the denominator. After that we can compute the limit.
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( { - 2 - h - 2x} \right)}}{h}$
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( { - 2 - h - 2x} \right)$
$ \Rightarrow f'\left( x \right) = - 2 - 2x$

Therefore, the derivative is, $f'\left( x \right) = - 2 - 2x$.

Note: We can directly find the derivative of given function using $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$, $\dfrac{d}{{dx}}\left[ {kf\left( x \right)} \right] = k\dfrac{d}{{dx}}f\left( x \right)$ and $\dfrac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$.
So, differentiating $f$ with respect to $x$.
$f'\left( x \right) = \dfrac{d}{{dx}}\left( {4 - 2x - {x^2}} \right)$
Now, use property $\dfrac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$.
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( 4 \right) - \dfrac{d}{{dx}}\left( {2x} \right) - \dfrac{d}{{dx}}\left( {{x^2}} \right)$
Now, use property $\dfrac{d}{{dx}}\left[ {kf\left( x \right)} \right] = k\dfrac{d}{{dx}}f\left( x \right)$.
$ \Rightarrow f'\left( x \right) = 4\dfrac{d}{{dx}}\left( {{x^0}} \right) - 2\dfrac{d}{{dx}}\left( x \right) - \dfrac{d}{{dx}}\left( {{x^2}} \right)$
Now, use property $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$.
$ \Rightarrow f'\left( x \right) = - 2 - 2x$
Therefore, the derivative is, $f'\left( x \right) = - 2 - 2x$.