Concept: Let small charge in x be ?x and the corresponding change in y is ?y.Therefore \(\rm ? y = \rm \frac{dy}{dx}? x\)Calculation:We have to find the value of \((64.02)^{1/6}\)Let x + ?x = 64.02 = 64 |+ 0.02Therefore, x = 64 and ?x = 0.02Assume, \(\rm y = x^{1/6}\) Differentiating with respect to x, we get\(\rm \Rightarrow \frac{dy}{dx} = \frac{1}{6}x^{-5/6} = \frac{1}{6(x)^{5/6}}\)At x = 64\(\rm \left[\frac{dy}{dx} \right ]_{x=64} = \frac{1}{192}\) and y = \((64)^{1/6} = 2\)As we know \(\rm ? y = \rm \frac{dy}{dx}? x\)So, \(\rm ? y = \frac{1}{192} \times (0.02) = 0.00010\)Therefore, approximate value of \((64.02)^{1/6}\) = y + ?y = 2 + 0.00010 = 2.00010