At which point is the tangent line to the cur | vedclass

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Question

(A) Given the curve $y = \cos(x + y)$.Taking the derivative with respect to $x$:$\frac{dy}{dx} = -\sin(x + y) \cdot (1 + \frac{dy}{dx})$$\frac{dy}{dx}

Vedclass · Accepted Answer

$(\pi/2, 0)$

Vedclass · Answer

(A) Given the curve $y = \cos(x + y)$.Taking the derivative with respect to $x$:$\frac{dy}{dx} = -\sin(x + y) \cdot (1 + \frac{dy}{dx})$$\frac{dy}{dx} (1 + \sin(x + y)) = -\sin(x + y)$$\frac{dy}{dx} = -\frac{\sin(x + y)}{1 + \sin(x + y)}$Since the tangent is parallel to $x + 2y = 0$,its slope is $-\frac{1}{2}$.Setting the derivative equal to $-\frac{1}{2}$:$-\frac{\sin(x + y)}{1 + \sin(x + y)} = -\frac{1}{2}$$2\sin(x + y) = 1 + \sin(x + y)$$\sin(x + y) = 1$This implies $x + y = \frac{\pi}{2} + 2n\pi$.Substituting this into the original equation $y = \cos(x + y)$:$y = \cos(\frac{\pi}{2} + 2n\pi) = 0$.Since $y = 0$,then $x + 0 = \frac{\pi}{2} \implies x = \frac{\pi}{2}$.Thus,the point is $(\frac{\pi}{2}, 0)$.

At which point is the tangent line to the curve $y = \cos(x + y)$,$x \in [-2\pi, 2\pi]$ parallel to the line $x + 2y = 0$?

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