# Thread: Chain rule partial derivative

1. ## Chain rule partial derivative

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

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

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

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

The value of $\, \dfrac{\partial g}{\partial t}\, (1,\, 1)\,$ 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!