91Ó°ÊÓ

The provided expression is $\mathrm{f}\left(\mathrm{x}\right)={∫}_0^{\mathrm{x}}\frac{{\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{t}}^2+0.51}\mathrm{dt}.$

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

Consider a function $\mathit{y}\mathbf{=}\mathit{f}\left(x\right)$, find the value of a function$\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$when$\mathit{x}\mathbf{=}\mathit{x}\mathbf{\text{'}}$.

As an example, the derivative of a function$\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$with respect to xwill be employed.

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

If the value of $x=x\text{'}$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 $y=f\left(x\right)$corresponding to the change dx in x. In practice, however, the concept of vanishingly small is not possible.

Use the expression for $f\left(x\right)$as an integral, apply the differential rules noting that $\frac{df}{dx}$will be the expression inside the integral evaluate at $t=x$.

$\mathrm{f}\left(\mathrm{x}\right)={∫}_0^{\mathrm{x}}\frac{{\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{t}}^2+0.51}\mathrm{dt}$

Since,

$df=\left(\frac{df}{dx}\right)dx$

Let x changes from 0.7 to 0.71.

$$\begin{gathered} dx=0.71=0.70 \\ =0.01 \end{gathered}$$

Putt all the provided values in above equation,

$\frac{df}{dx}=\frac{e^{-x}}{x^2+0.51}$

Here,

$$\begin{gathered} {\overline{)df=\frac{{\mathrm{e}}^{-x}}{x^2+0.51}}}_{x=0.7}^{}×0.01 \\ =\frac{e^{-0.7}}{0.7^2+0.51}×0.01 \end{gathered}$$

Therefore,

$df=0.00496≈0.005$