Question
Using differentials, find the approximate values of the following:
$\frac{1}{(2.002)^2}$

Answer

consider the function $\text{y}=\text{f} (\text{x})=\frac{1}{\text{x}^2}$

Let:

x = 2

$\text{x}+\triangle \text{x}=2.002$

Then,

$\triangle\text{x}=- 0.002$

For x = 2

$\text{y}=\frac{1} {2^2}=\frac{1}{4}$

Let:

$\text{dx}=\triangle \text{x}=0.002$

Now, $\text{y}=\frac{1} {\text{x}^2}$

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

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=2} =\frac{1}{4}$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1}{4}\times- 0.002=-0.005$

$\Rightarrow\triangle \text{y}=-0.0005$

$\therefore\frac{1} {(2.002)^2}=\text{y}+\triangle\text{y} =0.2495$

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