Answer:
Consider, \[y={{x}^{1/10}}.\] Let x = 1 and \[\Delta x=-\,0.001\] Then, \[\Delta y={{(x+\Delta x)}^{1/10}}-{{x}^{1/10}}\] \[\Rightarrow \] \[\Delta y={{(1-0.001)}^{1/10}}-{{(1)}^{1/10}}\] \[\Rightarrow \] \[{{(0.999)}^{1/10}}=1+\Delta y\] ? (i) Now, dy is approximately equal to \[\Delta y\] and is given by \[dy=\left( \frac{dy}{dx} \right)\cdot \Delta x=\frac{1}{10}{{x}^{\frac{-\,9}{10}}}\cdot (-\,0.001)\] \[=\frac{1}{10}{{(1)}^{\frac{-\,9}{10}}}(-\,0.001)=-\,0.0001\] \[\therefore \] \[{{(0.999)}^{1/10}}=1+dy\] [using Eq (i)] \[=1+(-\,0.0001)\] \[=1-\,0.0001=0.9999\] Thus, the approximate value of \[{{(0.999)}^{1/10}}\] is\[0.9999\].
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