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Thread: Chain rule partial derivative

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    Chain rule partial derivative

    (1 pt) Suppose that [tex]\,\chi(s,\, t)\, =\, -4s^2\, -\, 2t^2,\,y\,[/tex] a function of [tex]\,(s,\, t)\,[/tex] with [tex]\, y(1,\, 1)\, =\, 1\,[/tex] and [tex]\, \dfrac{\partial y}{\partial t}\, (1,\, 1)\, =\, -2.[/tex]

    Suppose that [tex]\, u\, =\, xy,\, v\,[/tex] a function of [tex]\, x,\, y\, [/tex] with [tex]\, \dfrac{\partial v}{\partial y}\, (-6,\, 1)\, =\,4.[/tex]

    Now suppose that [tex]\, f(s,\,t)\, =\, u(x(s,\, t),\, y(s,\, t))\, [/tex] and [tex]\, g(s,\, t)\, =\, v(x(s,\, t),\, y(s,\, t)).\, [/tex] You are given:

    . . . . .[tex]\dfrac{\partial f}{\partial s}\, (1,\, 1)\, =\, -32,\, [/tex]. . .[tex]\dfrac{\partial f}{\partial t}\, (1,\, 1)\, =\, 8,\,[/tex]. . .[tex]\dfrac{\partial g}{\partial s}\, (1,\, 1)\, =\, -16.[/tex]

    The value of [tex]\, \dfrac{\partial g}{\partial t}\, (1,\, 1)\, [/tex] must be:



    dv/dt=dv/dx*dx/dt+dv/dy*dy/dt
    dx/dt=-4t -> evaluate at (1,1) =-4
    dv/dt=-4dv/dx+4(-2)
    dv/dt=-4dv/dx-8

    How can I find the missing dv/dx in order to get a value for dv/dt? Thanks!
    Last edited by stapel; 11-04-2015 at 09:07 PM. Reason: Typing out the text contained in the graphic.

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