# 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:

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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