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Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (ii) maths textbook solution

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Answer: 0.208

Hint: here we use the formula

Given: (0.009)^{\frac{1}{3}}              

Solution: consider y=x^{\frac{1}{3}} \text { Let } x=0.009 \text { and } \Delta x=0.001

\Delta x=(x+\Delta x)^{\frac{1}{3}}=(0.009)^{\frac{1}{3}}-(0.008)^{\frac{1}{3}}(0.009)^{\frac{1}{3}}=0.2+\Delta y

Now, dy is approximately equal to \Delta y and is given by,

\begin{aligned} &d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \\\\ &=\frac{3}{3 \times 0.04}(0.001)=\frac{0.001}{0.12} \\\\ &=0.008 \end{aligned}

Hence, the approximate value of (0.004)^{\frac{1}{3}} i \text { is } 0.2+0.08=0.208

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