Concept: Let small change 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 \(\sqrt{36.01}\)Let x + ?x = 36.01 = 36 + 0.01Therefore, x = 36 and ?x = -0.01Assume, \(\rm y = x^{1/2}\) Differentiating with respect to x, we get\(\rm \Rightarrow \frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}\)At x = 36\(\rm \left[\frac{dy}{dx} \right ]_{x=36} = \frac{1}{12}\) and y = \(\rm (36)^{1/2} = 6\)As we know \(\rm ? y = \rm \frac{dy}{dx}? x\)So, \(\rm ? y = \frac{1}{12} \times (0.01) = 0.000833\)Therefore, approximate value of \(\sqrt{36.01} = (36.01)^{1/2}\) = y + ?y = 6 + 0.00083 = 6.00083