Use differential to approximate $\sqrt{36.6}$

The correct answer is 6.05

Let $f(x)=\sqrt{x}$ where $x=36$

Let $\Delta x=0.6$

∴ $f(x+ \Delta x) = \sqrt{x+ \Delta x}$ = $\sqrt{36+ 0.6}$


Now by definition, approximately we can write,

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



As $f(x)=\sqrt{x}$, $f'(x) = \dfrac{1}{2 \sqrt{x}}$ $= \dfrac{1}{2 \sqrt{36}}$ $=\dfrac{1}{12}$


Putting these values in (i), we get

$\dfrac{1}{12}$ = $\dfrac{\sqrt{36.6} - 6}{0.6}$

⇒ $\sqrt{36.6}$ = $\dfrac{0.6}{12} + 6$ = $6.05$