Second order partial derivatives and the chain rule

  • Thread starter Chantry
  • Start date
  • #1
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Homework Statement


http://www.math.wvu.edu/~hjlai/Teaching/Tip-Pdf/Tip3-27.pdf
Example 7.

Not this question in particular, but it shows what I'm talking about.

I understand how they get the first partial derivative, but I'm completely lost as how to take a second one.

I have tried using the product rule in combination with the chain rule and I never get anywhere near the correct result. I have a feeling this is going to be on my calculus exam, so I want to make sure I understand it properly.

I've been looking at this for hours and can't wrap my head around. Could someone maybe explain step by step how to go from d/dr (dw/dr) to the result they got dor d2w/dr2? (I realize these are equivalent).

Homework Equations



Basically filled this out above.

The Attempt at a Solution



Basically filled this out above.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
vela
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Let

[tex]g(x,y)=w_x\cos\theta + w_y\sin\theta[/tex]

Now calculate

[tex]\frac{\partial g}{\partial r} = \frac{\partial g}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial r}[/tex]

keeping in mind θ is being held constant.
 
  • #3
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Thank you!

It's funny how just one line of math made me finally understand it.
 

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