As in Problem 11, estimate $\sqrt[\mathbf{3}]{\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{05}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{98}{\mathbf{)}}^{\mathbf{2}}}$.

The provided expression is$\sqrt[3]{(2.05{)}^2+(1.98{)}^2}$ .

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

Consider a function $\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, find the value of a function$\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$whenlocalid="1659173701009" $\mathit{x}\mathbf{=}\mathit{x}\mathbf{\text{'}}$.

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

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

Provide the differential of the function $f(x,y)=\sqrt[3]{x^2+y^2}$, to get an estimate of the variation then we find the new value of f.$f(x,y)=\sqrt[3]{x^2+y^2}$

for$\sqrt[3]{(2.05{)}^2+(1.98{)}^2}$

Provided$x=2;dx=0.05,y=2;dy=-0.02$

On differentiate,

$$\begin{gathered} df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy \\ =\frac{2xdx}{3{\left(x^2+y^2\right)}^{{\displaystyle \frac{2}{3}}}}+\frac{2ydy}{3{\left(x^2+y^2\right)}^{\frac{2}{3}}} \end{gathered}$$

Substitute the provide values in above equation.

$$\begin{gathered} df=\frac{2}{3}\times \frac{2\times \left(0.05\right)-2\times 0.02}{8^{{\displaystyle \frac{2}{3}}}} \\ =\frac{2}{3}\times \frac{0.10-0.04}{4} \\ =\frac{2}{3}\times \frac{0.06}{4} \end{gathered}$$

Therefore,$df=0.01$

Then calculate,

$\begin{array}{l}{f}_n=f+df\\ =2+0.01\\ =2.01\end{array}$