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

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Answer: 1.96875

Hint : Here, we use the formula

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

Given: (15)^{\frac{1}{4}}


Let y=x^{\frac{1}{4}}=(16)^{\frac{1}{4}}=2

Where , x=16

\begin{aligned} &x+\Delta x=15 \\\\ &\Delta x=-1 \end{aligned}

Now, y=x^{\frac{1}{4}}

Differentiating w.r.t x

\frac{d y}{d x}=\frac{1}{4} x^{\frac{-3}{4}}

Using, \Delta y=\frac{d y}{d x} \Delta x=\frac{1}{4 x^{\frac{3}{4}}} \times-1

Putting the value of x

\Delta y=\frac{-1}{4(16)^{\frac{3}{4}}}=\frac{-1}{4 \times 2^{3}}=-0.03125

\begin{aligned} (15)^{\frac{1}{4}} &=y+\Delta y \\\\ &=2+(-0.03125) \\\\ &=1.96875 \end{aligned}


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