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 \(\rm (242)^{1/5}\)Let x + ?x = 242 = 243 - 1Therefore, x = 243 and ?x = - 1Assume, \(\rm y = x^{1/5}\) Differentiating with respect to x, we get\(\rm \Rightarrow \frac{dy}{dx} = \frac{1}{5}x^{-4/5} = \frac{1}{5(x)^{4/5}}\)At x = 243\(\rm \left[\frac{dy}{dx} \right ]_{x=243} = \frac{1}{405}\) and y = \((243)^{1/5} = 3\)As we know \(\rm ? y = \rm \frac{dy}{dx}? x\)So, \(\rm ? y = \frac{1}{405} \times (- 1) = 0.0024\)Therefore, approximate value of \(\rm (242)^{1/5}\) = y + ?y = 3 - 0.0024 = 2.997