Question

Using differentials, find the approximate value of the following:
$\sqrt{37}$

Answer 1

Let $y=f\left(x\right)=\sqrt{x}$

This gives $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$.

Now, we've,

$\Delta y=\frac{dy}{dx}\Delta x$

or, $\Delta y=\frac{1}{2\sqrt{x}}\Delta x$ [ For $y=\sqrt{x}$].......(1)

Now, let us take $x=36$ and $\Delta x=1$, then we've from (1),

$\Delta y=\frac{1}{12}.1$

or, $\Delta y=\frac{1}{12}$.

Again we've,

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

or, $\frac{1}{12}=\sqrt{37}-6$ [ For $x=36$ and $\Delta x=1$]

or, $\sqrt{37}=\frac{73}{12}$.