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

Answers (1)


Answer: 25^{\frac{1}{3}}=2.926

Hint: Here we use this below formula

        \Delta y=f(x+\Delta x)-f(x)

Given: 25^{\frac{1}{3}}

Solution: Let y=x^{\frac{1}{3}}

where x=27 \text { and } \Delta x=-2

Since y=x^{\frac{1}{3}}

        \begin{aligned} &\frac{d y}{d x}=\frac{d\left(x \frac{1}{3}\right)}{d x} \\\\ &=\frac{1}{3} x^{\frac{1}{3}-1} \\\\ &=\frac{1}{3} x^{\frac{-2}{3}}=\frac{1}{3 x^{\frac{2}{3}}} \end{aligned}


        \Delta y=\frac{d y}{d x} \Delta x

Putting values

        \begin{aligned} &\Delta y=\frac{1}{3(27)^{\frac{2}{3}}} \times(-2) \\\\ &\Delta y=\frac{-2}{3 \times\left(3^{3}\right)^{\frac{2}{3}}}=\frac{2}{3 \times 3^{2}}=\frac{2}{27}=-0.074 \end{aligned}

        \begin{aligned} &(25)^{\frac{1}{3}}=3-0.074 \\\\ &(25)^{\frac{1}{3}}=2.926 \end{aligned}

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