$\sqrt{36.6}$ is near $\sqrt{36}$ so we will use $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}$ with $\mathrm{x}=36$ and $∆\mathrm{x}=0.6$
Note that ${\mathit{f}}^{\prime }\left(\mathit{x}\right)=\frac{1}{2\sqrt{\mathit{x}}}$
$\sqrt{36.6}=\mathrm{f}\left(\mathrm{x}+∆\mathrm{x}\right)$
$\approx \mathrm{f}\left(\mathrm{x}\right)+{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)∆\mathrm{x}$
$=\sqrt{\mathrm{x}}+\frac{1}{2\sqrt{\mathrm{x}}}∆\mathrm{x}$
$=\sqrt{36}+\frac{1}{2\sqrt{36}}0.6$
$=6.05$
OR
The given curve is $\mathit{y}=2{\cos }^2\left(3\mathit{x}\right)$ .
Differentiating both sides with respect to $\mathrm{x}$ , we get:
$\frac{\mathrm{dy}}{\mathrm{dx}}=-12\mathit{s}\mathit{i}\mathit{n}\left(3\mathit{x}\right)\cos \left(3\mathit{x}\right)$
Now, at $\mathit{x}=\frac{\mathit{\pi }}{6}$ , we have:
$\frac{\mathrm{dy}}{\mathrm{dx}}=-12\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{2}\right)\cos \left(\frac{\mathit{\pi }}{2}\right)=0$
This means that the slope of tangent to the curve at $\mathrm{x}=\frac{\mathit{\pi }}{6}$ is zero.