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Need solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (x)

Answers (1)

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Answer: 1.004343

Hint: Here we use

        \Delta y=f(x+\Delta x)-f(x)
Given:

Solution:

let x=10

            x+\Delta x=10.1

Then

\begin{aligned} &\Delta x=0.1 \\\\ &\text { For } x=y \log _{10}=1 \\\\ &\text { Let: } d x=\Delta x=0.1 \end{aligned}\begin{aligned} &\Delta x=0.1 \\\\ &\text { For } x=y \log _{10}=1 \\\\ &\text { Let: } d x=\Delta x=0.1 \end{aligned}\begin{aligned} &\Delta x=0.1 \\\\ &\text { For } x=y \log _{10}=1 \\ &\text { Let: } d x=\Delta x=0.1 \end{aligned}

Now,   y=\log _{10} x=\frac{\log e x}{\log e 10}=\frac{1}{2.3025 x}

        \begin{aligned} &\left(\frac{d y}{d x}\right)_{x=10}=0.04343 \\\\ &\Delta y=d y=\frac{d y}{d x} x d x=0.04343 \times 0.1 \\\\ &\Delta y=0.004343 \end{aligned}        


\log _{10} 10.1=y+\Delta y=1.004343

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