Differentials, Errors and Approximations — Maths STD 12 Science — Question

$\triangle\text{x} =25.1-25$

$\triangle\text{x} =0.1$

Let $\text{y}=\frac{1} {\sqrt{\text{x}}}$

$\frac{\text{dy}} {\text{dx}}=\frac{2}{\text{x}^{\frac{3} {2}}}$

$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}=25}=-\frac{1} {2(25)^{\frac{3}{2}}}$

$=-\frac{1}{250}$

$=-0.004$

Now,

$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=25}\times(\triangle\text{x})$

$=(-0.004)(0.1)$

$=-0.0004$

$\frac{1}{\sqrt{25.1}} =\text{y}+\triangle\text{y}$

$=\frac{1}{\sqrt{\text {x}}}+(-0.0004)$

$=\frac{1}{\sqrt{25}}- 0.0004$

$=\frac{1}{5}-0.0004$

$=0.2-0.0004$

$\frac{1}{\sqrt{25.1}} =0.1996$