Find the equation of the tangent to the curve y = -cotx at the point (pi/4, -1)

Show any derivatives that you need to find when solving this problem

Question

Find the equation of the tangent to the curve y = -cotx at the point (pi/4, -1)

Show any derivatives that you need to find when solving this problem

anonymous · Answer

I know that -cotx differentiates to cosec2x but idk how to progress further

anonymous · Answer

Develop an equation for a straight line using point-slope form:

$$(y-y1) = m(x-x1)$$
where (x1,y1) = (pi/4,-1) and m = y'(pi/4)

anonymous · Answer

How do i differentiate cosec^2x and (pi/4)?

anonymous · Answer

$$y \prime (x) = - csc^2(x)$$
Solve for y'(pi/4)

anonymous · Answer

You dont differentiate again

anonymous · Answer

Ok so how do you differentiate (pi/4) though?

anonymous · Answer

The answer says 2 but idk how they got that

anonymous · Answer

You plug in pi/4 for x in the above equation then use for the straight line equation

anonymous · Answer

ok thanks, I got the final answer now

anonymous · Answer

y = 2x-2.57