Show that the approximate relative error $\frac{\left(df\right)}{\mathbf{f}}$ of a product$\mathit{f}\mathbf{=}\mathit{g}\mathit{h}$ is the sum of the approximate relative errors of the factors.

The given equations are $\frac{df}{f}$ and$f=gh$.

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$\mathit{d}\mathit{x}\mathbf{\Rightarrow }\mathbf{0}$ .

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

Consider the relative errors for two variables, g and h , as well as h , $g_{re}$, and $h_{re}$. As a result, it can use the specified relation to estimate the relative error in$f\left(f_{re}\right)$ .

It must, however, first acquire the differentials of the relationship:

Since,

$f=gh$

Taking log on both sides,

$\mathrm{In}f=\mathrm{In}g+\mathrm{In}h$

Here,

$\frac{df}{f}=\frac{dg}{g}+\frac{dh}{h}$

Hence,$f_{re}=g_{re}+h_{re}$