Y=-ctg(x+π/3) M(0;-√3)

Предыдущий
вопрос Следующий
вопрос

To find the equation of the tangent line to the curve y=-ctg(x+π/3) at the point M(0, -√3), we need to find the derivative of the function y=-ctg(x+π/3) and then evaluate it at x=0 to find the slope of the tangent line.

Find the derivative of y=-ctg(x+π/3)
y = -ctg(x+π/3
Take the derivative with respect to x
y' = -c * (-csc^2(x+π/3))

Evaluate the derivative at x=0
When x=0, csc(π/3) = 2/√
y' = -c (-2/√3)^
y' = -c 4/
y' = -4c/3

Find the slope of the tangent line at M(0, -√3)
The tangent line's slope is equal to the derivative evaluated at x=0
m = -4c/3

The equation of the tangent line
m = -4c/
Given point M(0, -√3), substitute into the point-slope form equation y-y1 = m(x-x1)
y-(-√3) = (-4c/3)(x-0
y+√3 = (-4c/3)
y = (-4c/3)x - √3

Therefore, the equation of the tangent line to the curve y=-ctg(x+π/3) at the point M(0, -√3) is y = (-4c/3)x - √3.