Question

Use differentials to find the approximate value of $\sqrt{36.1}$ (for four decimal places)

Answer 1

Let $y=f\left(x\right)=x^{\frac{1}{2}}$
Let $x=36,\Delta x=0.1$
Thus, $y=x^{\frac{1}{2}}$
$\therefore \Delta y=\frac{1}{2}{x}^{-\frac{1}{2}}\Delta x$
Here $x=36,\Delta x=0.1$
Therefore, $\Delta y=\frac{1}{2}{\left(36\right)}^{-\frac{1}{2}}\left(0.1\right)$
$\Rightarrow \frac{1}{2}\left(\frac{1}{6}\right)\left(0.1\right)=0.0083$
$\Rightarrow f(x+\Delta x)=y+\Delta y=\sqrt{36}+\Delta y$
$\Rightarrow 6+0.0083=6.0083$