#### Need solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (xxii)

Hint: Here we use this below formula

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

Given: $\left(\frac{17}{81}\right)^{\frac{1}{4}}$

Solution: Let  $y=x \frac{1}{4}$

where $x=16 \text { and } \Delta x=1$

Since   $y=x^ \frac{1}{4}$

$\frac{d y}{d x}=\frac{d\left(x ^\frac{1}{4}\right)}{d x}=\frac{1}{4} x^{\frac{1}{4}-1}=\frac{1}{4} x^{\frac{-3}{4}}=\frac{1}{4 x^{\frac{3}{4}}}$

\begin{aligned} &\Rightarrow \text { Now, } \Delta y=\frac{d y}{d x} \Delta x=\frac{1}{4} \times \frac{1}{16^{\frac{3}{4}}} \times 1 \\\\ &=\frac{1}{4} \times \frac{1}{24^{\frac{3}{4}}}=\frac{1}{4} \times \frac{1}{2^{3}}=\frac{1}{32} \end{aligned}

Now,

$(17)^{\frac{1}{4}}=y+\Delta y$

Putting values,

\begin{aligned} &(17)^{\frac{1}{4}}=(16)^{\frac{1}{4}}+\Delta y \\\\ &(17)^{\frac{1}{4}}=(24)^{\frac{1}{4}}+\Delta y \end{aligned}

\begin{aligned} &(17)^{\frac{1}{4}}=2+\frac{1}{32} \\ &(17)^{\frac{1}{4}}=2.03125 \end{aligned}

Now,

$\left(\frac{17}{81}\right)^{\frac{1}{4}}=\frac{2.03125}{3}=0.677$