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I know this is probably quite basic but having trouble with the following question:

Let f be a function of x and y, where x = cos(uv) and y = sin(uv).

Use the chain rule to show that:

$\displaystyle \frac{\partial f}{\partial u} = v(x\frac{\partial f}{\partial y}-y\frac{\partial f}{\partial x})$

The problem I have is I get the following:

$\displaystyle \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$          (1)

$\displaystyle \frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$          (2)

$\therefore$ from (1):

$\displaystyle \frac{\partial f}{\partial u}=\frac{\partial x}{\partial u}(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial v}\frac{\partial v}{\partial x})$      (3)

&

$\displaystyle \frac{\partial f}{\partial v}=\frac{\partial y}{\partial v}(\frac{\partial f}{\partial y} - \frac{\partial f}{\partial u}\frac{\partial u}{\partial y})$      (4)

Then:

$\displaystyle \frac{\partial x}{\partial u} = -vsin(uv), \frac{\partial x}{\partial v} = -usin(uv), \frac{\partial y}{\partial u} = vcos(uv), \frac{\partial y}{\partial v} = ucos(uv)$

Plugging in the partial derivatives for x and y in to (3) I get:

$\displaystyle \frac{\partial f}{\partial u}=-vy(\frac{\partial f}{\partial x} + \frac{1}{uy}\frac{\partial f}{\partial v})$

Doing the same with (4) I get:

$\displaystyle \frac{\partial f}{\partial v}=ux(\frac{\partial f}{\partial y} - \frac{1}{vx}\frac{\partial f}{\partial u})$

After substituting for $\displaystyle \frac{\partial f}{\partial v}$ and some algebra, I get the following:

$\displaystyle \frac{\partial f}{\partial u}=-vy\frac{\partial f}{\partial x} - vx\frac{\partial f}{\partial y} + \frac{\partial f}{\partial u}$

So I'm close but not quite there. Any thoughts? What silly mistake have I made?

Thanks in advance.

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1 Answer 1

up vote 1 down vote accepted

I think that you only have to use the chain rule with $x, y$ being the intermediate variables, thus getting

$$ \frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}$$

and then since

$$\frac{\partial x}{\partial u} = -v \sin{(uv)} = -v y$$

and

$$\frac{\partial y}{\partial u} = v \cos{(uv)} = v x$$

you should get what you want

$$\frac{\partial f}{\partial u} = -vy \frac{\partial f}{\partial x} + vx \frac{\partial f}{\partial y} = v \left ( x \frac{\partial f}{\partial y} - y \frac{\partial f}{\partial x} \right )$$

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I just clearly over complicate these things. Thanks Adrián. –  Billy Ray Valentine Apr 6 '12 at 7:56
    
Thinking about it further, u is a function of x and y hence it's the sum of the partial derivatives with respect to x and y, using the chain rule. This was really helpful and it should hopefully stop me from over complicating things in the future. –  Billy Ray Valentine Apr 7 '12 at 21:33

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