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Chain rule partial derivative

wololo

New member
(1 pt) Suppose that \(\displaystyle \,\chi(s,\, t)\, =\, -4s^2\, -\, 2t^2,\,y\,\) a function of \(\displaystyle \,(s,\, t)\,\) with \(\displaystyle \, y(1,\, 1)\, =\, 1\,\) and \(\displaystyle \, \dfrac{\partial y}{\partial t}\, (1,\, 1)\, =\, -2.\)

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

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

. . . . .\(\displaystyle \dfrac{\partial f}{\partial s}\, (1,\, 1)\, =\, -32,\, \). . .\(\displaystyle \dfrac{\partial f}{\partial t}\, (1,\, 1)\, =\, 8,\,\). . .\(\displaystyle \dfrac{\partial g}{\partial s}\, (1,\, 1)\, =\, -16.\)

The value of \(\displaystyle \, \dfrac{\partial g}{\partial t}\, (1,\, 1)\, \) must be:

[HR][/HR]
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!
 
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