Partial Derivative

[Sandeep Kumar]

Could someone please help me solve this?

If f(x,y) is a function of two variables x and y which are themselves function of other two variables p and q where x= p(q^2), y= p+ (1/q) show that:

 

[Steven G]

Sure we can help you solve this problem. Can you please tell us where you are stuck so that we can explain what you need to do next? Please post the work which you have done so far?

 

[topsquark]

Perhaps you should start by writing out the extra step
[math]\dfrac{\partial f}{\partial p} = \dfrac{ \partial f}{ \partial x} \cdot \dfrac{ \partial x}{ \partial p} + \dfrac{ \partial f}{ \partial y} \cdot \dfrac{ \partial y}{ \partial p}[/math]
Now use [math]x = pq^2[/math] and [math]y = p + \dfrac{1}{q}[/math] to find your [math]\dfrac{ \partial x}{ \partial p}[/math] and [math]\dfrac{ \partial y}{ \partial p}[/math]. (Or note the comparison between the two lines.)

For the second partials Is this a problem with the concept or is it a problem with the calculation? Otherwise do the same thing as in part 1. I would hate to have to write [math]\dfrac{ \partial f}{ \partial p \partial q} = \dfrac{ \partial \left ( \dfrac{ \partial f}{ \partial p} \right ) }{ \partial q}[/math] too many times, but you can simply write it out term by term. Yes, it's ugly, but normally in any school work you would know f(p, q) anyway.

-Dan

 

[HallsofIvy]

Do you not know the "chain rule" for functions of more than one variable? If f is a function of x and y and x and y are themselves functions of p and q then
\(\displaystyle \frac{\partial f}{\partial p}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial p}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial p}\)
and
\(\displaystyle \frac{\partial f}{\partial q}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial q}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial q}\)

Since you are given x and y in terms of p and q, you can immediately find \(\displaystyle \frac{\partial x}{\partial p}\), \(\displaystyle \frac{\partial x}{\partial q}\), \(\displaystyle \frac{\partial y}{\partial p}\), and \(\displaystyle \frac{\partial y}{\partial q}\) and put those in.

 

[Steven G]

Perhaps you should start by writing out the extra step
[math]\dfrac{\partial f}{\partial p} = \dfrac{ \partial f}{ \partial x} \cdot \dfrac{ \partial x}{ \partial p} + \dfrac{ \partial f}{ \partial y} \cdot \dfrac{ \partial y}{ \partial p}[/math]
Now use [math]x = pq^2[/math] and [math]y = p + \dfrac{1}{q}[/math] to find your [math]\dfrac{ \partial x}{ \partial p}[/math] and [math]\dfrac{ \partial y}{ \partial p}[/math]. (Or note the comparison between the two lines.)

For the second partials Is this a problem with the concept or is it a problem with the calculation? Otherwise do the same thing as in part 1. I would hate to have to write [math]\dfrac{ \partial f}{ \partial p \partial q} = \dfrac{ \partial \left ( \dfrac{ \partial f}{ \partial p} \right ) }{ \partial q}[/math] too many times, but you can simply write it out term by term. Yes, it's ugly, but normally in any school work you would know f(p, q) anyway.

-Dan

Dan, I missed your point about Or note the comparison between the two lines. Can you please explain what you mean? Thanks!