#### explain solution RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 4 maths

Hint: area of a cubical box of radius $x$ is given by $s=6 x^{2}$

Given: $\Delta x$ is the error in the value of $x \; \; \Delta x=0.01 x$

Solution: suppose $x$ be the of cubical box

So,$\Delta x=0.01 x$

Differentiate  $s$ with respect to $x$

\begin{aligned} &\frac{d s}{d x}=\frac{d}{d x}\left(6 x^{2}\right) \\\\ &\Rightarrow \frac{d s}{d x}=6 \times \frac{d}{d x}\left(x^{2}\right) \\\\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \Rightarrow \frac{d s}{d x}=6 x 2 x=12 x \end{aligned}

⇒ As we know that if   $y=F(x) \Delta x$  is a small increment in $x$ , then the corresponding increment in  $y$

$\Delta x=F(x+\Delta y)-F(x)$  , is approximately given as  $\Delta y=\left(\frac{d y}{d x}\right) \Delta x$

\begin{aligned} &\Rightarrow \text { Here, } \frac{d s}{d x}=12 x \Delta x=0.01 x \\\\ &\Delta s=12 x \times 0.01 x \\\\ &\Delta s=0.12 x^{2} \end{aligned}

$\Rightarrow$  Percentage of error is

$\text { Error }=\frac{0.12 x^{2}}{6 x^{2}} \times 100 \%=0.02 \times 100 \%=2 \%$