Question

If $ x $ changes from 4 to 4.01, then find the approximate change in $ {\log _e}x $

Answer 1

Hint: We can use the formula of first principle method to find the change in $ {\log _e}x $ . Finding the derivative of a function by computing the limits is known as differentiation from first principles.

Complete step-by-step answer:
Let $ f(x) = $ $ {\log _e}x $
We know that $ f'(x) = \dfrac{{\lim }}{{\Delta x \to 0}}\dfrac{{f(x + \Delta x) - f(x)}}{{\Delta x}} $
If $ \Delta x $ is not tending to zero but is a very small increment. Then we can remove limit and write
 $ f'(x) = \dfrac{{f(x + \Delta x) - f(x)}}{{\Delta x}} $
 $ \Rightarrow \Delta xf'(x) = f(x + \Delta x) - f(x) $ . . . . . (1)
Where $ \Delta x $ is a small increment $ f(x) $ is value of the function at $ x $
Then, $ f(x + \Delta x) - f(x) $ represents the change in $ f(x) $ when $ x $ changes from $ x $ to $ x + \Delta x $ .
Now, $ f(x) = {\log _e}x $
By differentiating with respect to $ x $ , we get
 $ f'(x) = \dfrac{1}{x} $ $ \left( {\because \dfrac{d}{{dx}}{{\log }_{_e}}x = \dfrac{1}{x}} \right) $
 $ \Rightarrow {(f'(x))_{x = 4}} = \dfrac{1}{4} $
From equation (1), we can write
Change in $ {\log _e}x = f(x + \Delta x) - f(x) $
 $ = \Delta xf'(x) $ . . . (2)
 $ \Delta x = 4.01 - 4 = 0.01 $
 $ \therefore $ change in $ {\log _e}x $ $ = 0.01 \times \dfrac{1}{4} $ [from equation (2)]
 $ = \dfrac{{0.01}}{4} $
 $ \Rightarrow \Delta x = 0.0025 $
Hence, the approximate change in $ {\log _e}x $ is 0.0025.

Note: Such type of questions cannot be solved just by mugging up the formulae. You need to understand the concept behind it to make sure you never do it wrong. In simple terms, limit is approximation and derivative is a small increment in the original value.