## Calculus: Early Transcendentals (2nd Edition)

Taking a=144, $\sqrt (146)\approx \frac{145}{12}$
Take y= $\sqrt a$. Let a=144 and let ∆x= 2 Then ∆y= $\sqrt (a+∆x)-\sqrt a= \sqrt 146-\sqrt 144= \sqrt 146- 12$ Or $\sqrt 146= ∆y+12$ Now dy is approximately equal to ∆y and is given by, dy= $(\frac{dy}{dx})∆x=\frac{1}{2\sqrt a}(2)= \frac{1}{\sqrt 144}=\frac{1}{12}$ Thus the approximate value of $\sqrt 146$ is $12+\frac{1}{12}= \frac{145}{12}$