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

Answers (1)

Answer: 4.042

Hint: here we use the formula

        \Delta y=f(x+\Delta x)-f(x)

Given: 66^{\frac{1}{3}}

Solution:   Consider the function y=f(x)=x^{\frac{1}{3}}

Let,

\begin{aligned} &x=64 \\\\ &x+\Delta x=66 \end{aligned}

Then,

\begin{aligned} &\Delta x=2 \\\\ &\Rightarrow \text { for } x=64 \\\\ &y=(64)^{\frac{1}{3}}=4 \end{aligned}

Let,

\begin{aligned} &d x=\Delta x=2 \\\\ &\text { Now, } y=\left(x^{\frac{1}{3}}\right) \\\\ &\frac{d y}{d x}=\frac{1}{3(x)^{\frac{2}{3}}} \\\\ &\left(\frac{d y}{d x}\right)=\frac{1}{48} \end{aligned}

\begin{aligned} &\Delta y=d y=\frac{d y}{d x} d x=\frac{1}{48} \times 2=0.042 \\\\ &\Delta y=0.042 \\\\ &(66)^{\frac{1}{3}}=y+\Delta y=4.042 \end{aligned}

 

 

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