Partial derivative chain rule

**sgcb** · September 29th 2012, 07:34 AM

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 $g_{v}(0, 0)$ .

	$f$	$g$	$f_{x}$	$f_{y}$
(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?

**HallsofIvy** · September 29th 2012, 08:53 AM

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 $g_{v}(0, 0)$ .

	$f$	$g$	$f_{x}$	$f_{y}$
(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 $f_y$ .

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

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

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