# Partial derivative chain rule

• Sep 29th 2012, 07:34 AM
sgcb
Partial derivative chain rule
Q: Suppose $f$ is a differentiable function of $x$ and $y$, and $g(u, v) = f(e^u + sin(v), e^u + cos(v))$. Use the table of values to calculate $g_{u}(0, 0)$ and $g_{v}(0, 0)$.

 img.top {vertical-align:15%;} $f$ img.top {vertical-align:15%;} $g$ img.top {vertical-align:15%;} $f_{x}$ img.top {vertical-align:15%;} $f_{y}$ (0, 0) 3 6 4 8 (1, 2) 6 3 2 5

I apply the chain rule for 2 independent/2 dependent variable how I /think/ it's suppose to be applied in this problem:

$\frac{\partial x}{\partial u}=e^u$, $\frac{\partial x}{\partial v}=cos(v)$

$\frac{\partial y}{\partial u}=e^u$, $\frac{\partial y}{\partial v}=-sin(v)$

$\frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial u} = 4*e^u + 8*e^u = 12$

$\frac{\partial g}{\partial v}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial v} = 4cos(v) - 8sin(v) = 4$

But the answer in BOB is 7, 2... Where is my error in understanding?
• Sep 29th 2012, 08:53 AM
HallsofIvy
Re: Partial derivative chain rule
Quote:

Originally Posted by sgcb
Q: Suppose $f$ is a differentiable function of $x$ and $y$, and $g(u, v) = f(e^u + sin(v), e^u + cos(v))$. Use the table of values to calculate $g_{u}(0, 0)$ and $g_{v}(0, 0)$.

 img.top {vertical-align:15%;} $f$ img.top {vertical-align:15%;} $g$ img.top {vertical-align:15%;} $f_{x}$ img.top {vertical-align:15%;} $f_{y}$ (0, 0) 3 6 4 8 (1, 2) 6 3 2 5

I apply the chain rule for 2 independent/2 dependent variable how I /think/ it's suppose to be applied in this problem:

$\frac{\partial x}{\partial u}=e^u$, $\frac{\partial x}{\partial v}=cos(v)$

$\frac{\partial y}{\partial u}=e^u$, $\frac{\partial y}{\partial v}=-sin(v)$

$\frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial u} = 4*e^u + 8*e^u = 12$

You are saying that $f_x= 4$ and $f_y= 8$
Those are the values given in your table (nicely done, by the way) for (0, 0). But you are asked to find the values at u= 0, v= 0. Because $x= e^u+ sin(v)$ and $y= e^u+ cos(v)$, when u= v= 0, $x= e^0+ sin(0)= 1$ and $y=e^0+ cos(0)= 2$. you need to use the (1, 2) row to find $f_x$ and $f_y$.

Quote:

$\frac{\partial g}{\partial v}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial v} = 4cos(v) - 8sin(v) = 4$

But the answer in BOB is 7, 2... Where is my error in understanding?