Solve the problem.
-Find the tangent to $y = \cot x$ at $x = \frac { \pi } { 4 }$.
A) $y = - 2 x + \frac { \pi } { 2 }$
B) $y = 2 x - \frac { \pi } { 2 } + 1$
C) $y = - 2 x + \frac { \pi } { 2 } + 1$
D) $y = 2 x + 1$

Question

Quizplus · Accepted Answer

First, we need to find the derivative of $y = \cot x$.
$$\frac{dy}{dx} = -\csc^2 x$$
At $x = \frac{\pi}{4}$, $\csc(\frac{\pi}{4}) = \sqrt{2}$.
So, the slope of the tangent line at $x = \frac{\pi}{4}$ is 
$$\left. \frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = -(\sqrt{2})^2 = -2$$
Next, we need to find the equation of the tangent line using point-slope form. 
Using the point $(\frac{\pi}{4}, \cot(\frac{\pi}{4})) = (\frac{\pi}{4}, 1)$, we have
$$y-1 = -2(x-\frac{\pi}{4})$$
Simplifying the equation, we get
$$y = -2x + \frac{\pi}{2} + 1$$
Therefore, the answer is C.

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