Use differential to approximate $\sqrt{8.95}$

The correct answer is 2.99

Let $f(x)=\sqrt{x}$ where $x=9$

Let $\Delta x=-0.05$

∴ $f(x+ \Delta x) = \sqrt{x+ \Delta x}$ = $\sqrt{9 - 0.05}$


Now by definition, approximately we can write,

$f'(x) = \dfrac{f(x+ \Delta x) -f(x)}{\Delta x}$ ----(i)



As $f(x)=\sqrt{x}$, $f'(x) = \dfrac{1}{2 \sqrt{x}}$ $= \dfrac{1}{2 \sqrt{9}}$ $=\dfrac{1}{6}$


Putting these values in (i), we get

$\dfrac{1}{6}$ = $\dfrac{\sqrt{8.95} - \sqrt{9}}{-0.05}$

⇒ $\sqrt{8.95}$ = $\dfrac{-0.05}{6} + 3$ = $2.99$