Answer

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**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.

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