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The problem is as follows:

The question

My working is:

Page 1 Page 2

I don't know what I should be doing. Can you please help?

ahorn
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1 Answers1

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According to your first line,

$g(x,y)=f(u(x,y),v(x,y))$.

This does not imply $\frac{\partial g}{\partial x}=\frac{\partial f}{\partial x}$ (or "similarly for $y$").

In line 2, I believe what you are trying to say is $$ \frac{\partial g}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x} $$.

Subject to these changes in interpreting the chain rule and the role of variables for partial derivatives, you are not far from executing the right logical steps to obtain the required answer.

You can avoid the problem of notation by working directly with respect to the correct variable:

$$g_x(x,y)=f_x(x+y,-x+y) \cdot \frac{\partial}{\partial x}(x+y)+f_y(x+y,-x+y) \cdot \frac{\ \partial}{\partial x} (-x+y)$$

$$g_y(x,y)=f_x(x+y,-x+y) \cdot \frac{\partial}{\partial y}(x+y)+f_y(x+y,-x+y) \cdot \frac{\ \partial}{\partial y} (-x+y)$$

which leads to Option (c).

OnceUponACrinoid
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  • Why can't I use g and f interchangeably? They are equal to each other. Using your method, I still get $\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}=2\frac{\partial f}{\partial u}$ and the RHS does not correspond to any of the choices. – ahorn Jun 06 '15 at 19:57
  • Option 3 leads to the same answer. The notation $\frac{\partial f}{\partial u}$ is referring to the derivative of $f$ with respect to the first variable. I believe your problem may be from interpreting the notation. Note that to someone using $f$ who does not know the definition of $g$, the variable $u$ does not really exist or make sense. – OnceUponACrinoid Jun 06 '15 at 20:10
  • I do not understand how option (c) is the same. Can you please show me what steps to do to convert $2\frac{\partial f}{\partial u}$ into the RHS of option (c)? – ahorn Jun 06 '15 at 21:09
  • This is a notation issue. Suppose $g(x,y)=2y$. To write $g(x,y)=f(x+y,-x+y)=f(u,v)$, the appropriate $f(u,v)$ would equal $u+v$. Note that $\frac{\partial{f}}{\partial u}=1$. This is the same as saying $\frac{\partial f}{ \partial x}=1$ if $f(x,y)=x+y$. Please take a look at the edited answer. – OnceUponACrinoid Jun 06 '15 at 21:24
  • Did you substitute $u=x$ and $y=v$ to get $f_u(x,y)=f_x(u,v)$? – ahorn Jun 07 '15 at 14:22
  • Not really. If you look at the last two lines in the answer above -- you can see that the chain rule is being used without reference to $u$ or $v$. Maybe you can look at some multi-variable chain rule example to get familiar with the notation. – OnceUponACrinoid Jun 07 '15 at 16:07