Question

Using differential, find the approximate value of the following:
$\sqrt{26}$

Answer 1

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

Now, $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$.

Now,

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

or, $\Delta y=\frac{1}{2\sqrt{x}}\Delta x$

Now for $x=25$ and $\Delta x=1$ we've,

$\Delta y=\frac{1}{2\sqrt{25}}\times 1=0.1$......(1)

Again,

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

or, $\Delta y=\sqrt{26}-\sqrt{25}$ [ for $x=25$ and $\Delta x=1$]

or, $\sqrt{26}=5+0.1$ [ Using (1)]

or, $\sqrt{26}=5.1$.