Question

Using differentials, the approximate value of $(26{)}^{1/3}$ is $2.9629$

• A. True
• B. False

Answer 1

The correct option is A True

Let $y=x^{\left(\frac{1}{3}\right)}x=27,\Delta x=-1y=3,\Delta y=\left(\frac{dy}{dx}\right)\times \Delta x\left(\frac{dy}{dx}\right)=+\left(\frac{1}{3}\right)x^{\left(\frac{-2}{3}\right)}=\left(\frac{1}{3}\right)\times \left(27{)}^{(}\frac{-2}{3}\right)=\left(\frac{1}{27}\right)\therefore \Delta y=\left(\frac{1}{27}\right)\times -1=\left(\frac{-1}{27}\right)\therefore newy=y+\Delta y=3-\left(\frac{1}{27}\right)2.9629$