Use differentials to show that, for large n and small a, $\sqrt{\mathbf{n}\mathbf{+}\mathbf{a}}\mathbf{-}\sqrt{\mathbf{n}}\mathbf{\cong }\frac{\mathbf{a}}{\mathbf{2}\sqrt{\mathbf{n}}}$.Find the approximate value of $\sqrt{{\mathbf{10}}^{\mathbf{26}}\mathbf{+}\mathbf{5}}\mathbf{-}\sqrt{{\mathbf{10}}^{\mathbf{26}}}$.

As the provided equation is $\sqrt{{10}^{26}+5}-\sqrt{{10}^{26}}$

And to prove $\sqrt{n+a}-\sqrt{n}\cong \frac{a}{2\sqrt{n}}$.

This method is based on the derivatives of functions whose values must be calculated at certain locations, as the name implies.

Consider a function y = f (x), find the value of a function y = f (x)when x =x'.

As an example, the derivative of a function y = f (x)with respect to xwill be employed.

$\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{=}\mathbf{\left(}\mathbf{f}\left(x\right)\mathbf{\right)}$is Change in y with respect to change in xas dx$\to$0.

If the value of x =x' from a value of x near it, such that the difference in the two values, dx, is vanishingly small, one can derive the change in the value of the function y = f(x) corresponding to the change dx in x. In practice, however, the concept of vanishingly small is not possible.

The stated equation is $\frac{1}{{\left(n+1\right)}^3}-\frac{1}{n^3}\cong \frac{3}{n^4}$.

If $f\left(x\right)=\frac{1}{x^3}$is defined, the necessary differences are $∆f=f\left(n+1\right)-f\left(n\right)$. However, because$∆f$ is close to $∆df$(for extremely large n),

the above differences can be rewritten using differentials:

$$\begin{gathered} df=\frac{df}{dx}dx \\ =d\left(\sqrt{x}\right) \\ =\frac{1}{2}\frac{1}{\sqrt{x}}dx \end{gathered}$$

with x = n and dx = a

$$\begin{gathered} df=\frac{1}{2}\frac{1}{\sqrt{x}}\times a \\ =\frac{a}{2\sqrt{n}} \end{gathered}$$

Now, provided$\sqrt{{10}^{26}+5}-\sqrt{{10}^{26}}$

Here,$n={10}^{26};a=5$

Then,$\frac{a}{2\sqrt{n}}=\frac{5}{2\sqrt{{10}^{26}}}\cong 2.5\times {10}^{-13}$

Hence,$\sqrt{{10}^{26}+5}-\sqrt{{10}^{26}}\cong 2.5\times {10}^{-13}$