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

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentials, Errors and Approximations4 Marks
Question
Using differentials, find the approximate values of the following:
$\log_\text{e}10.01$ it being given that $\log_{10}=0.4343$
✓

Answer

Consider the function $\text{y}=\text{f} (\text{x})=\log_{10}\text{x}$

Let:

x = 10

$\text{x}+\triangle \text{x}=10.01$

Then,

$\triangle\text{x} =0.01$

For x

$\text{y}\log_{10} 10=1$

Let:

$\text{dx}=\triangle \text{x}=0.01$

Now, $\text{y}=\log_ {10}\text{x}=\frac{\log_\text{e}\text {x}}{\log_\text{e}10}$

$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1} {2.3025\text{x}}$

$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =10}=0.04343$

$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx} =0.04343\times0.01=0.004343$

$\Rightarrow\text{y} =0.004343$

$\therefore\log_{10} 10.01=\text{y}+\triangle\text{y} =1.004343$

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