# B Chain rule problem

1. Feb 26, 2017

### knockout_artist

(Sorry for the mistakes first thread using hand held device)

Hello,
I was working on Harold T. Davis
Introduction to Nonlinear Differential and Integral Equations

I saw this following equations

1-So equation 4 came as a result of chain rule applies on equation. 3 ?

2- how did equation 5 come about? And specially the "2" in second term.

3-What kind of calculus I should study, so I know this sort of stuff and I can save some time.

Thank you.

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2. Feb 26, 2017

### Simon Bridge

The point of this sort of book is for you to explore how you understand calculus

1-So equation 4 came as a result of chain rule applies on equation. 3 ?

... you can answer this for yourself: what happens when you apply the chain rule to eq3?

2- how did equation 5 come about? And specially the "2" in second term.
... have you tried just differentiating eq4 wrt x? What happens when you do?

3-What kind of calculus I should study, so I know this sort of stuff and I can save some time.
... differential calculus. "Differential equations" is the usually subject heading from 2nd year college.

3. Feb 26, 2017

### knockout_artist

3.with product rule I get

do I need to take Y' as well and add it?

4. Feb 26, 2017

### Simon Bridge

I do not know what you are trying to show... the "3" could refer to my point 3, which has nothing to do with the product rule... or it could refer to eq3, which does not get to your pic using the product rule.

5. Feb 27, 2017

### knockout_artist

I actually tried to differentiating eq4 wrt x. I got this(from your point 2). Sorry about the confusion.

I learned ODE/PDE from Keryzsig's "advance engineering mathematics".
How ever I also have Tenenbaum/Pollard's "ordinary differential equation", which does not have such topic covered.

6. Feb 27, 2017

### PeroK

One thing you are missing (and the book is treating implicitly) is that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are functions of two variables.

7. Feb 27, 2017

### knockout_artist

Now I get this

even if I add them I don't get equation 5.

8. Feb 27, 2017

### PeroK

That looks close. Aren't you just missing that last term in $y_{xx}$?

Just on my phone so hard to see exactly what you are missing!

9. Feb 28, 2017

### PeroK

Here's a breakdown in shorthand notation, starting with:

$(f_x) + (f_y)( y_x)$

And differentiating with respect to $x$ gives:

$(f_{xx} + f_{xy} y_x) + (f_{yx} + f_{yy} y_x)(y_x) + (f_y)(y_{xx}) = f_{xx} + 2f_{xy} y_x + f_{yy} (y_x)^2 + f_y y_{xx}$

10. Feb 28, 2017

### knockout_artist

On LHS term in the middle seems like differentiated wrt to y, rather then x.

11. Feb 28, 2017

### PeroK

Which term? Remember that $f_x$ and $f_y$ are functions of two variables, so the chain rule applies to them as it does to $f$ in the first place.

12. Mar 1, 2017

### knockout_artist

I see that.
Thank you very much for your help.