1. Using differentials, find the approximate value of each of the following up to 3 places of decimal.

(iii)  \sqrt {0.6}
  

Answers (1)

Lets suppose y = \sqrt x and let x = 1 and \Delta x = -0.4
Then,
\Delta y = \sqrt{x+\Delta x} - \sqrt x
\Delta y = \sqrt{1+(-0.4)} - \sqrt 1
\Delta y = \sqrt{0.6} - 1
\sqrt{0.6} = \Delta y +1
Now, we cam say that \Delta y  is approximately equals to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(-0.4) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = -0.4)\\ dy = \frac{1}{2\sqrt 1}.(-0.4)\\ dy = \frac{1}{2}.(-0.4)\\ dy = -0.2
Now,
\sqrt{0.6} = \Delta y +1\\ \sqrt {0.6} = (-0.2) + 1\\ \sqrt{0.6} = 0.8
Hence, \sqrt{0.6} is approximately equal to 0.8



        

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