How do you use linear approximation to the Square root function to estimate square root $\sqrt {4.400} $ ?

Hint: As per this question, we have to find the square root by using linear approximation to the square root function. Consider a value closer to the given value and find the square root of that value. Find the first derivative of the closer value and substitute the values in linear approximation formula to find the square root of the given function.

Formula used: The linear approximation of f at ‘$a$ ’ is $L(x) = f(a) + f'(a)(x - a)$

Complete step-by-step solution:
Square root of a number $x$ is that number which when multiplied by itself gives the number $x$ itself is a perfect square.
Here, the function we need to approximate is
$f(x) = \sqrt x $
The value of $x$ is $4.400$
$\therefore $ We want to estimate $f(4.400) = \sqrt {4.400} $
To find that we need to find a value closer to $4.400$ which is the value of ‘$a$ ’
So, let us take closer number as $4$
Now, $f(a) = \sqrt a $
Substitute the value of $a$
$ \Rightarrow f(4) = \sqrt 4 $
The square root of $4$ is $2$
$\therefore f(4) = 2$
Let us find the derivative of $f(a)$ with respect to $a$
 $ \Rightarrow f'(a) = \dfrac{d}{{da}}{(a)^{\dfrac{1}{2}}}$
When we differentiate this using $\left( {\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$
We get,
\[ \Rightarrow f'(a) = \dfrac{1}{2}{a^{\left( {\dfrac{1}{2} - 1} \right)}}\]
By reducing the above equation,
\[ \Rightarrow f'(a) = \dfrac{1}{2}{a^{ - \dfrac{1}{2}}}\]
We can also write the equation as
 $f'(a) = \dfrac{1}{{2\sqrt a }}$
Substituting the value of $a$ in above equation, we get
 $f'(4) = \dfrac{1}{4}$
Now using the linear approximation
 $L(x) = f(a) + f'(a)(x - a)$
Replacing the values
 $ \Rightarrow L\left( x \right) = f\left( 4 \right) + f'\left( 4 \right)\left( {4.400 - 4} \right)$
Substitute the values of $f\left( 4 \right)$ and$f'\left( 4 \right)$,
 $ \Rightarrow L\left( x \right) = 2 + \dfrac{1}{4}\left( {4.400 - 4} \right)$
 $\therefore $The value of $\sqrt {4.400} = 2.1$

Therefore the value of $\sqrt {4.400}$ is 2.1.

Note: Linear approximation, or linearization, is a method we can use to approximate the value of a function at a particular point.
The linear approximation is useful because it can be difficult to find the value of a function at a particular point.
Square roots are a great example of this.
The symbol of the square root is $\sqrt {} $ . For example the square root of a number $4$ is represented as $\sqrt 4 = 2$