Question

Using differentials, find approximate value of $(0.37{)}^{1/2}$.

Answer 1

$\sqrt{0.37}$

Let $y=\sqrt{x}$

$x=0.36$ and $\Delta x=0.01$

$\because y=\sqrt{x}$

Differentiating above equation, we have

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

Now,

$\Delta y=\frac{dy}{dx}\left(\Delta x\right)$

$\Rightarrow \Delta y$ $=\frac{1}{2\sqrt{x}}\left(0.01\right)=\frac{1}{2\sqrt{0.36}}\left(0.01\right)=\frac{1}{2\times 0.6}\left(0.01\right)=\frac{0.01}{1.2}$

Also,

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

$\Rightarrow \Delta y=\sqrt{x+\Delta x}-\sqrt{x}$

$\Rightarrow \frac{0.01}{1.2}=\sqrt{0.37}-\sqrt{0.36}$

$\Rightarrow \sqrt{0.37}=\frac{0.01}{1.2}+0.6=\frac{0.01+0.72}{1.2}=\frac{0.73}{1.2}=0.608$

Hence the value of $\sqrt{0.37}$ is $0.608$.