Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 172: 5

0.003333

$ f(x)=\sqrt {x+6}$, $ a=3$, $\Delta x=0.02$ In order to find $ f'(x)$, we let $ u=x+6$. Then, $ f(x)=y=\sqrt u $ and $ f'(x)=\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\frac{1}{2\sqrt u}\times1=\frac{1}{2\sqrt {x+6}}$ Recall: $ f(a+\Delta x)-f(a)\approx f'(a)\Delta x $ $\implies f(3+0.02)-f(3)\approx f'(3)\Delta x=\frac{1}{2\sqrt {3+6}}\times0.02=0.003333$