How do you find the difference quotient of $f$, that is, find $\left( {\dfrac{{f(x + h) - f(x)}}{h}} \right),h \ne 0$ for $f(x) = {x^2} - 5x + 7$?

Hint: if you understand the question correctly, you have to start by substituting $(x + h)$ wherever you see $x$ in your original function given i.e. $f(x) = {x^2} - 5x + 7$ and then simplify the equation so obtained after substitution to get the desired answer.

Complete step by step solution:
It is given in the question that,
$f(x) = {x^2} - 5x + 7$
Now replace $x$by $(x + h)$ which gives us
$ \Rightarrow {(x + h)^2} - 5(x + h) + 7$
On multiplying, you have
$ \Rightarrow {x^2} + 2hx + {h^2} - 5(x + h) + 7$
$ \Rightarrow {x^2} + 2hx + {h^2} - 5x - 5h + 7$
So, substitute the value of $f(x + h)$ in the definition of the difference quotient.
$\left( {\dfrac{{f(x + h) - f(x)}}{h}} \right),h \ne 0$
$\therefore $ $\left( {\dfrac{{f(x + h) - f(x)}}{h}} \right) = \dfrac{{({x^2} + 2hx + {h^2} - 5x - 5h + 7) - ({x^2} - 5x + 7)}}{h}$
On simplifying we get
$\left( {\dfrac{{f(x + h) - f(x)}}{h}} \right) = \dfrac{{{x^2} + 2hx + {h^2} - 5x - 5h + 7 - {x^2} + 5x - 7}}{h}$
On grouping similar terms and solving them we get,
$\left( {\dfrac{{f(x + h) - f(x)}}{h}} \right) = \dfrac{{2hx + {h^2} - 5h}}{h}$
Now , since this is calculus, the next step is to find the limit of the function where $h \to 0$ .
For this, we cannot have h in the denominator because h approaches 0.
Therefore, taking h common from both numerator and denominator and simplifying we get,
\[ \Rightarrow \dfrac{{h(2x - 5 + h)}}{{h(1)}}\]
$ \Rightarrow 2x + h - 5$
Put h=0, in the above equation we get
$ \Rightarrow 2x - 5$
Which is nothing but the derivative of the original function $f(x) = {x^2} - 5x + 7$

Note:
Differentiation can be defined as a derivative of independent variable value and can be used to calculate features in an independent variable per unit modification.
Let,
$y = f(x)$ be a function of $x$ .
Then, the rate of change of per unit change in is given by,
$\dfrac{{dy}}{{dx}}$
If the function, $f(x)$ undergoes an infinitesimal change of h near to any point $x$, then the derivative of the function is depicted as ,
$\mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{f(x + h) - f(x)}}{h}} \right)$
When a function is depicted as $y = f(x)$,
Then the derivative is depicted by the following notation:
$D(y)$ or $D[f(x)]$ is called the Euler’s notation.
$\dfrac{{dy}}{{dx}}$ is known as Leibniz’s notation.
$F'(x)$ is known as Lagrange’s notation.
Differentiation is the method of evaluating a function’s derivative at any time.