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# Using differentials the approximate value of $\sqrt {401}$ is(A) $20.100$ (B) $20.025$ (C) $20.030$ (D) $20.125$

Last updated date: 03rd Aug 2024
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Hint: Use the formula of differential for approximation. You can derive that formula using the mathematical expression of the first principle method by removing the limit sign and replacing $h$ with $\Delta x$ . Once you have that formula, a separate 401 is such a way that you can find the square root of one part and name the second part as $\Delta x$ . Then substitute these values in the formula.

The formula of differentiation by first principle method is given by
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
We can remove the limit sign by replacing $h$ with $\Delta x$ we can write
$f'(x) = \dfrac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$
By cross multiplying, we get
$\Delta xf'(x) = f(x + \Delta x) - f(x)$
By rearranging it, we can write
$f(x + \Delta x) = f(x) + \Delta xf'(x)$ . . . (1)
This is the formula for approximation using differential, in which
$x + \Delta x$ is the total value given whose approximation we have to calculate
$x$ is the value whose definite answer we know
$\Delta x$ is the remaining value, also called increment or decrement. We choose it depending upon the given value.
Now, for this example, since, we know the square root of 400. We will write
$x = 400$
$\Delta x = 1$
$f(x) = \sqrt x = \sqrt {400}$
$\Rightarrow f'(x) = \dfrac{1}{{2\sqrt x }}$ $\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$
$f(x + \Delta x) = f(401) = \sqrt {401}$
Therefore, from equation (1), we get
$f(401) = f(x) + \Delta xf'(x)$
$= \sqrt {400} + 1 \times \dfrac{1}{{2\sqrt {400} }}$
$= 20 + \dfrac{1}{{2 \times 20}}$
$= 20 + \dfrac{1}{{40}}$
$= 20 + 0.025$
$\Rightarrow \sqrt {401} = 20.025$
Therefore, from the above explanation, the correct answer is, option (B) $20.025$
So, the correct answer is “Option B”.

Note: You do not need to remember the derivation of the formula of approximation. But it is better to know how it came. The key point in approximation is choosing $\Delta x$ . You have to choose it in such a way that you can get a definite value of $f(x)$ .