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

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Answer: 20.025

Hint: Here, we use the formula

        \Delta y=f(x+\Delta x)-f(x)

Given: \sqrt{401}

Solution

Let  y=x^{\frac{1}{2}}

Where, x=400

\begin{aligned} &x+\Delta x=401 \\\\ &\Delta x=1 \end{aligned}

Now, \frac{d y}{d x}=\frac{1}{2 \sqrt{x}}

Using,  \Delta y=\frac{d y}{d x} \Delta x=\frac{1}{2 \sqrt{x}} \times 1

Putting the value of x

\begin{aligned} \sqrt{401} &=y+\Delta y \\\\ &=20+0.025 \\\\ &=20.025 \end{aligned}

 

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