How do you use linear approximation to find the value of $ {\left( {1.01} \right)^{10}} $ ?

Hint: In this question, we have to find the value of $ {\left( {1.01} \right)^{10}} $ using the linear approximation method. In order to apply this method, we need to use the formula $ y = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) $ for a function $ f\left( x \right) $ which we will take as $ f\left( x \right) = {x^{10}} $ and for such a point $ a $ which is close to the value we need to find, thus $ a = 1 $ . We will put the $ x = 1.01 $ in the formula to obtain the value of $ y $ which would be our answer.

Complete step-by-step answer:
(i)
We are asked to find the approximate value of $ {\left( {1.01} \right)^{10}} $ through a linear approximation method. In this method we will apply the formula:
 $ y = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) $
Where, $ f\left( x \right) $ is the function we will define by looking at the exponent we have got in our question and $ a $ will be the approximated and nearest value of $ x $ .
Therefore, $ f\left( a \right) $ will be the value of the function $ f\left( x \right) $ at the point $ a $ and $ f'\left( a \right) $ will be the value of the derivative of the function $ f\left( x \right) $ at the point $ a $ .
As we are asked to calculate $ {\left( {1.01} \right)^{10}} $ , we can clearly define our function $ f\left( x \right) $ as $ {x^{10}} $ . Therefore,
 $ f\left( x \right) = {x^{10}} $
Since, we need the derivative of the function too, we will differentiate both sides with respect to $ x $ . Therefore, we will get:
 $ f'\left( x \right) = 10{x^9} $
(ii)
In the function $ f\left( x \right) = {x^{10}} $ , we want the value of $ x $ to be $ 0.01 $ in order to obtain our answer and since the value of $ a $ will be the approximated value of $ x $ , we will have $ a = 1 $ .
Therefore, the values of $ f\left( a \right) $ and $ f'\left( a \right) $ will be:
 $ f\left( a \right) = f\left( 1 \right) = {1^{10}} = 1 $
And,
 $ f'\left( a \right) = f'\left( 1 \right) = 10 \times {1^9} = 10 $
(iii)
Now, since we have all the values to put in the formula, we will put $ x = 1.01 $ to obtain the value of $ y $ which would be our answer. Therefore,
 $
  y = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) \\
  y = 1 + 10\left( {1.01 - 1} \right) \\
  y = 1 + 10\left( {0.01} \right) \\
  y = 1 + 0.1 \\
  y = 1.1 \;
  $
Since, we get $ y = 1.1 $ , we can say that through linear approximation method, $ {\left( {1.01} \right)^{10}} = 1.1 $
So, the correct answer is “1.1”.

Note: Linear approximation is also known as linearization. It is a method to approximate the value of a function at a particular point. It is a quick and simple method that estimates a value otherwise it is very difficult to find. But note that since it is a rough approximation, the further away you get from the point $ x = a $ , the less precise the approximation becomes.