Question

\(y=|\cos x|+|\sin x|\) \(y=\sin x-\cos x\) is the neighbourhood of \(x=\frac{2 \pi}{3}\) \(y^{\prime}=\cos x+\sin x\) \(y^{\prime}=\frac{-1}{2}+\frac{\sqrt{3}}{2}\)

Short answer

Answer: The value of the second function at \(x=\frac{2\pi}{3}\) is \(\frac{1}{2}+\frac{\sqrt{3}}{2}\), and the slope of the function at this point is \(\frac{-1}{2}+\frac{\sqrt{3}}{2}\).

Step-by-step solution

Evaluate the functions at \(x=\frac{2\pi}{3}\)

We first need to evaluate both functions at the given point \(x=\frac{2\pi}{3}\): $$y_1=|\cos\left(\frac{2\pi}{3}\right)|+|\sin\left(\frac{2\pi}{3}\right)|$$ $$y_2=\sin\left(\frac{2\pi}{3}\right)-\cos\left(\frac{2\pi}{3}\right)$$

Simplify and confirm equality

Simplify both expressions: $$y_1=|-\frac{1}{2}|+|\frac{\sqrt{3}}{2}|=\frac{1}{2}+\frac{\sqrt{3}}{2}$$ $$y_2=\frac{\sqrt{3}}{2}-\left(-\frac{1}{2}\right)=\frac{\sqrt{3}}{2}+\frac{1}{2}$$ Since both \(y_1\) and \(y_2\) evaluate to the same value, we can now proceed to compute the derivative of the second function.

Compute derivative of \(y=\sin x-\cos x\)

Differentiate the second function to find the slope of the function. $$y^{\prime}=\frac{d}{dx}\left(\sin x-\cos x\right) = \cos x+\sin x$$

Evaluate the derivative at \(x=\frac{2\pi}{3}\)

Substitute the given value of \(x=\frac{2\pi}{3}\) into the expression for the derivative: $$y^{\prime}\left(\frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)+\sin\left(\frac{2\pi}{3}\right)$$ $$y^{\prime}\left(\frac{2\pi}{3}\right) = \frac{-1}{2}+\frac{\sqrt{3}}{2}$$ The slope of the function \(y=\sin x-\cos x\) at \(x=\frac{2\pi}{3}\) is \(\frac{-1}{2}+\frac{\sqrt{3}}{2}\).