Using differentials, Find the approximate value of $(66{)}^{1/3}$.

Consider ${66}^{\frac{1}{3}}$$y=f\left(x\right)=x^{\frac{1}{3}}$

We need to find approximate value of ${66}^{\frac{1}{3}}$.

Let $x=64$ and $x+∆x=66$

Then ,$∆x=66-x=66-64=2$.

For $x=64$,

Then, $y=f\left(64\right)={64}^{\frac{1}{3}}={\left(4^3\right)}^{\frac{1}{3}}=4$

As $dx=∆x \Rightarrow dx=2$

By differentiating $y=x^{\frac{1}{3}}$, we get

$\frac{dy}{dx}=\frac{1}{3}{x}^{\left(\frac{1}{3}-1\right)}=\frac{1}{3}{x}^{-\frac{2}{3}}=\frac{1}{3{\left(x\right)}^{{\displaystyle \frac{2}{3}}}}$

As the formula of differentiation of $x^n=n{\left(x\right)}^{n-1}$

Now, $$\begin{gathered} {\left(\frac{dy}{dx}\right)}_{x=64}=\frac{1}{3{\left(64\right)}^{{\displaystyle \frac{2}{3}}}}=\frac{1}{3{\left(4^3\right)}^{{\displaystyle \frac{2}{3}}}} \\ =\frac{1}{3{\left(4\right)}^2}=\frac{1}{3\times 16}=\frac{1}{48} \end{gathered}$$

We know $∆y=dy=\frac{dy}{dx}\left(dx\right)$

By substituting $\frac{dy}{dx} , dx$ in above equation, we get

$∆y=\frac{1}{48}\times 2=\frac{1}{24}$$=0.041666$$\cong 0.042$

As $66=x+∆x,$ then ${\left(66\right)}^{\frac{1}{3}}=y+∆y$

By substituting $y,∆y$in above equation 

$\Rightarrow y+∆y=4+0.042=4.042$

Therefore , the approximate value of ${66}^{\frac{1}{3}}=4.042$.