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