Question

Use differential to approximate $\sqrt{36.6}$.

Answer 1

$\sqrt{36.6}$ is near $\sqrt{36}$ so we will use $f\left(x\right)=\sqrt{x}$ with $x=36$ and $\Delta x=0.6$

Note that $f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$

$\sqrt{36.6}=f(x+\Delta x)\approx f\left(x\right)+f^{\prime }\left(x\right)\Delta x=\sqrt{x}+\frac{1}{2\sqrt{x}}\Delta x=\sqrt{36}+\frac{1}{2\sqrt{36}}0.6=6.05$