## Calculus (3rd Edition)

$f(x)=\sqrt {x+6}$, $a=3$, $\Delta x=0.02$ 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: $f(a+\Delta x)-f(a)\approx f'(a)\Delta x$ $\implies f(3+0.02)-f(3)\approx f'(3)\Delta x=\frac{1}{2\sqrt {3+6}}\times0.02=0.003333$