If x = cot(y) what is dy/dx?

Here we will use:

cot(x) = cos(x)/sin(x)

cosec2(x) = 1 + cot2(x)

Chain rule : dy/dt * dt/ dx = dy/ dx

Product rule : d/dx (uv) = udv/dx + v*du/dx

x = cot(y)= cos(y)/sin(y)

dx/dy = -cos(y)(cos(y)/sin(y)2) + 1/sin(y) (-sin(y))

         = -cos2(y)/sin2(y) -1 = - cos2(y)/sin2(y) -sin2(y)​ /sin2(y)

         = -(sin2(y)+ cos2(y))/sin2(y)​ = -cosec2(y)

cosec2(y) = 1 + cot2(y)

x = cot(y)

dx/dy = -(1 + x2)

dy/dx = -1/(1+x2)

SS
Answered by Sahiti S. Maths tutor

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