Use implicit differentiation to show

dy/dx = [(1-xy)cot(y)]/[x^2 cot(y) + xcsc^2(y)]

if xy = ln(x cot y).

1 answer

  1. from xy = ln(x coty)
    x dy/dx + y = (x(-csc^y dy/dx) + coty)/(x coty)
    cross-multiply
    (x^2)(coty)dy/dx + xycoty = -x(csc^2y)dy/dx + coty
    (x^2)(coty)dy/dx + x(csc^2y)dy/dx = coty - xycoty
    dy/dx((x^2)(coty) + x(csc^2y)) = coty(1 - xy)
    dy/dx = (1-xy)coty/[(x^2)(coty)dy/dx + x(csc^2y)]

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