Using differentials, find the approximate value of $(25)^{\dfrac{1}{3}}$ upto 3 places of decimal.

The correct answer is $2.926$

Let $y=x^{\dfrac{1}{3}}$ where $x=27$ and $\Delta x= -2$


∴ $ \Delta y$ = $(x+ \Delta x)^{\dfrac{1}{3}}$ - $(x)^{\dfrac{1}{3}}$ = $(25)^{\dfrac{1}{3}} - 3$



As $y=x^{\dfrac{1}{3}}$, $\dfrac{dy}{dx} = \dfrac{1}{3(x)^\dfrac{2}{3}}$


Now by definition, approximately we can write,


$\dfrac{dy}{dx}$ = $\dfrac{(y+ \Delta y) - y}{\Delta x}$ ⇒ $\Delta y$ = $\dfrac{dy}{dx}$ $\Delta x$ ----(i)



Putting values in (i), we get


$(25)^{\dfrac{1}{3}} - 3$ = $\dfrac{1}{3(x)^\dfrac{2}{3}}$ $\times (-2)$


⇒ $(25)^{\dfrac{1}{3}}$ = $\dfrac{1}{3(27)^\dfrac{2}{3}}$ $\times (-2) + 3$ = $3 - \dfrac{2}{27}$ = $2.926$