# Partial derivative chain rule

• Dec 11th 2008, 01:38 PM
DavidWagner
Partial derivative chain rule
Let z = g(x, y) and suppose that x(t) = t^2 + 3t + 2 and y(t) = e^t + sin(3t)

Find (dz/dt)|t=0

if (dg/dx)|(1,2) = 6
(dg/dy)|(1,2) = -2
(dg/dx)|(2,1) = -3
(dg/dy)|(2,1) = 8
(dg/dx)|(0,0) = 0
(dg/dy)|(0,0) = -4

Thank You.
• Dec 11th 2008, 02:02 PM
TheEmptySet
Quote:

Originally Posted by DavidWagner
Let z = g(x, y) and suppose that x(t) = t^2 + 3t + 2 and y(t) = e^t + sin(3t)

Find (dz/dt)|t=0

if (dg/dx)|(1,2) = 6
(dg/dy)|(1,2) = -2
(dg/dx)|(2,1) = -3
(dg/dy)|(2,1) = 8
(dg/dx)|(0,0) = 0
(dg/dy)|(0,0) = -4

Thank You.

By the Chain rule we get

$\displaystyle \frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} +\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial g}{\partial x}(2t+3)+\frac{\partial g}{\partial y}(e^t+3\cos(3t))$

Now we need to do some calculations

$\displaystyle x(0)=2$ and $\displaystyle y(0)=1$

so $\displaystyle \frac{\partial g}{\partial x}\bigg |_{(2,1)}=-3$
and
$\displaystyle \frac{\partial g}{\partial y} \bigg |_{(2,1)}=8$

so Finally we get

$\displaystyle \frac{\partial z}{\partial t}=(-3)(2(0)+3)+8(e^0+3\cos(0))=-9+32=23$