# I don't understand these 2 equations from my textbook?

• Apr 11th 2013, 08:10 AM
HowDoIMath
I don't understand these 2 equations from my textbook?
Attachment 27906

For this one, when I try differentiate y(x) I end up with dy/dx = (2xy)(dy/dx) + y^2 instead of dy/dx = 2xy (I'm assuming y(x) and y mean virtually the same thing?)

Attachment 27907
For this one I tried to verify that this example was correct, but I found that dM/dy = 2(dx/dy) + cosy + (e^-x)(y)(dx/dy)-e^-x
and dN/dx = (-xsiny)(dy/dx) + cosy - (siny)(dy/dx) - e^-x.
So it seems that the terms with (dy/dx) or (dx/dy) automatically become zeroes or just don't matter, but I don't understand why?
• Apr 11th 2013, 08:41 AM
HallsofIvy
Re: I don't understand these 2 equations from my textbook?
I don't know where you got that " $y^2$" from. "Leibniz formula" says
$\frac{d}{dx}\int_\alpha(x)^\beta(x) f(x,t) dt= \frac{d\beta}{dx}f(x, \beta(x))- \frac{d\alpha}{dx}f(x,\alpha(x))+ \int_\alpha(x)^\beta(x)\frac{\partial f(x, t)}{\partial x} dt$

Here, $\alpha(x)= 1$ so its derivative is 0, and $\beta(x)= x$ so its derivative is 1. $f(x,t)= ty(t)$. That does not depend upon x so [tex]\frac{\partial f(x, t)}{\partial x}= 0[tex]
That is the derivative of the integral is simply f(x,x)= 2xy(x).

As for the second, you differentiate with respect to x and y independently, that is, you take the partial derivatives. There should be no "dy/dx" in the formulas.
• Apr 11th 2013, 02:08 PM
HowDoIMath
Re: I don't understand these 2 equations from my textbook?
Attachment 27908
Here's what I got when I tried to derive the first equation. I never learned the Leibniz formula so this is pretty confusing to me.

For the 2nd one, I never learned about partial derivatives and I didn't notice the different "d" symbol used. Thanks for your help on this I understand the 2nd part now.

Note: I don't know why there's this attached thumbnail, but I can't get rid of it.
• Apr 11th 2013, 04:50 PM
HallsofIvy
Re: I don't understand these 2 equations from my textbook?
You are integrating incorrectly. $\int_0^x ty(t)dt$ is NOT $\frac{t}{2}y^2(t)$. For one thing you are not integrating the "t", for another y is a function of t, not just "t" itself.
For example, $\int sin(t)dt$ is NOT $sin^2(t)$.
• Apr 11th 2013, 06:21 PM
HowDoIMath
Re: I don't understand these 2 equations from my textbook?
Thank you very much.
Quote:

For one thing you are not integrating the "t", for another y is a function of t, not just "t" itself.
I think you mean "not just "y" itself?
• Apr 11th 2013, 06:34 PM
HallsofIvy
Re: I don't understand these 2 equations from my textbook?
Actually I meant "not just t itself". I don't know what you mean by "just y itself" since the integral is with respect to t. $\int t dt= \frac{1}{2}t^2+ C$. If y is a constant then $\int y dt= yt+ C$ and if y is a function of t, the integral depends on exactly what function it is.
• Apr 13th 2013, 03:56 PM
HowDoIMath
Re: I don't understand these 2 equations from my textbook?
oh wow I don't know how I got confused before... I'm just stuck in the habit of seeing y as a variable and not as a function. I've seen f(x) and y, but somehow seeing y(x) confused me here. For some y(t) the derivative would just be d(y(t)), which likely won't require the power rule when evaluated.