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11 April, 14:43

Find dy/dx by implicit differentiation: e^y cos (x) = 1 + sin (xy)

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Answers (2)
  1. 11 April, 15:06
    0
    Hello,

    f (x, y) = e^y cos (x) - 1-sin (xy) = 0

    @f/@x=-sin x * e^y-y*cos (xy)

    @f/@y=cosx * e^y-x*sin (xy)

    dy/dx = - (@f/@x) / (@f/@y) = (sin x * e^y+y*cos (xy)) / (cos x * e^y-x*sin (xy)
  2. 11 April, 17:37
    0
    Implicit expression refers to equation that are not strictly expressed in terms of y and x separately. In this case, the derivative of expression

    e^y cos (x) = 1 + sin (xy) is

    -e^ y sin x dx + cos x e^y dy = cos (xy) (xdy + y dx)

    dx (-e^ y sin x - y cosxy) = dy (x cos xy - cos x e^y)

    dy/dx = (-e^ y sin x - y cosxy) / (x cos xy - cos x e^y)
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