Use implicit differentiation to find dy/dx?

x + cot(xy) = -2

dy/dx = ?

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  • 8 years ago
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    dx+ csec^2(xy).( xdy + ydx)=0

    dy(x csec^2(xy)=(-1-ycsec^2(xy))dx

    dy/dx=(-1-ycsec^2(xy))/(x csec^2(xy)

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  • Josh
    Lv 6
    8 years ago

    There is an operator that might make this easier for you its the upper case D.

    D(x) means to differentiate x. Usually you would subscript the D with whatever variable you are differentiating with respect to. You will use this operator a lot if you take multivariate calculus. It is a very convenient way of ordering up multiple chain rules with out the clutter of the dy/dx stuff.

    x + cot(xy) = -2

    D(x + cot(xy) = D(-d) differentiate both sides

    D(X) + D(cot(xy)) = D(-2)

    1 + D(cot(xy)) = 0 simplify the easy parts

    1 + -csc^2(xy)D(xy) = 0 apply the chain rule

    1 + -csc^2(xy)*(D(x)y + xD(y)) = 0 apply the chain rule again

    1 + -csc^2(xy)(y + xy') = 0

    1 - ycsc^2(xy) - xy'csc^2(xy) = 0

    1 - ycsc^2(xy) = xy'csc^2(xy)

    (1-ycsc^2(xy)/(xcsc^2(xy) = y' solve for y'

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  • x + cot(xy) = -2

    d/dx [x + cot(xy)] = d/dx [-2]

    d/dx [x] + d/dx [cot(xy)] = 0

    1 - (csc(xy))^2 * d/dx [xy] = 0

    1 - (csc(xy))^2 * [x *dy/dx + y] = 0

    1 - x (csc(xy))^2 (dy/dx) - y (csc(xy))^2 = 0

    -x (csc(xy))^2 (dy/dx) = y(csc(xy))^2 - 1

    dy/dx = (y (csc(xy))^2 - 1) / (-x (csc(xy))^2)

    dy/dx = -y/x + (1/x) (sin(xy))^2

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  • 8 years ago

    x + Cot xy = - 2

    dy/dx = 1 - Csc² xy

                ¯¯¯¯¯¯¯¯¯

                  or

    dy/dx = 1 - (1 / Sin² xy)

                ¯¯¯¯¯¯¯¯¯¯¯¯

                  or

    dy/dx = (Sin² xy - 1) / Sin² xy

                ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

     

    Source(s): 10/22/12
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