Question

$\begin{aligned} &\mathrm{y}=\sqrt{3+2[\sin x \sin 2 \mathrm{x}+\sin x \sin 3 x+\sin 2 x \sin 3 x}\\\ &+\cos x \cos 2 x+\cos x \cos 3 x+\cos 2 x \cos 3 x]\\\ &=\sqrt{3+2[\cos x+\cos 2 x+\cos x]}\\\ &=\sqrt{3+4 \cos x+4 \cos ^{2} x-2}\\\ &=\sqrt{4 \cos ^{2} x+4 \cos x+1}\\\ &y=|2 \cos x+1|\\\ &\text { At } x=\frac{\pi}{2}\\\ &y=2 \cos x+1\\\ &y^{\prime}=-2 \sin x\\\ &\mathrm{y}^{\prime}=-2 \end{aligned}$ $\begin{aligned} &\text { At } x=\pi / 5 \\ &y=2 \cos x+1 \\ &y^{\prime}=-2 \sin x \\ &y^{\prime}=-2 \sin \pi / 5=\frac{\sqrt{5}+3}{2} \end{aligned}$

Short answer

Question: Find the derivative y' of the function \(y = |2 \cos x + 1|\) and evaluate y' at \(x = \frac{\pi}{2}\) and \(x = \frac{\pi}{5}\). Answer: The derivative of y with respect to x is \(y' = -2 \sin x\). When \(x = \frac{\pi}{2}\), \(y' = -2\), and when \(x = \frac{\pi}{5}\), \(y' = -2 \sin \frac{\pi}{5} = \frac{\sqrt{5} + 3}{2}\).

Step-by-step solution

Simplify the expression for y

Notice the given expression inside the square root can be rearranged as sum of products of trigonometric terms. Use the product-to-sum identities: \(\sin a \sin b = \frac{1}{2}(\cos(a - b) - \cos(a + b))\) and \(\cos a \cos b = \frac{1}{2}(\cos(a - b) + \cos(a + b))\) to simplify the expression: $\begin{aligned} y &= \sqrt{3+2[\sin x \sin 2x+\sin x \sin 3x+\sin 2x \sin 3x + \cos x \cos 2x+\cos x \cos 3x+\cos 2x \cos 3x]} \\ &= \sqrt{3+2[\frac{1}{2}(\cos x - \cos 3x)+\frac{1}{2}(\cos 2x - \cos 4x)+\frac{1}{2}(\cos x + \cos 5x) + \frac{1}{2}(\cos x + \cos 3x)+\frac{1}{2}(\cos 2x + \cos 4x)+\frac{1}{2}(\cos x - \cos 5x)]} \\ \end{aligned}$

Combine terms inside the square root and simplify further

Combine the terms inside the square root and cancel out any opposing terms: \(y=\sqrt{3+2[\frac{1}{2}(\cos x- \cos 3x+\cos 2x -\cos 4x+\cos x+\cos 5x +\cos x+\cos 3x+\cos 2x+\cos 4x+\cos x-\cos 5x)]}\) \(= \sqrt{3+2[2 \cos x+2 \cos 2x]}\)

Simplify the square root

Finish simplifying the square root: \(y=\sqrt{3+4[\cos x+\cos 2x]}\) \(=\sqrt{4 \cos^2 x+4 \cos x +1}\) \(=|2 \cos x+1|\)

Find the derivative of y with respect to x

Differentiate y with respect to x using the chain rule for the absolute value function and the derivative of the cosine function: \(y'=-2 \sin x\)

Evaluate y' at x = pi/2

Evaluate y' at \(x=\frac{\pi}{2}\): \(y'(-2\sin\frac{\pi}{2})=-2\)

Evaluate y' at x = pi/5

Evaluate y' at \(x=\frac{\pi}{5}\): \(y'(-2\sin\frac{\pi}{5})=-2 \sin \frac{\pi}{5}=\frac{\sqrt{5}+3}{2}\)