Can anyone please verify/confirm these derivatives?

In summary, @Math100's guess is correct: y is a function of t. @Math100 also confirmed that y' means ##\frac{dy}{dt}##.
  • #1
Math100
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Homework Statement
If ## F(x, y')=\sqrt{x^2+y'^2} ##, find ## \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial y'}, \frac{dF}{dx} ## and ## \frac{d}{dx}(\frac{\partial F}{\partial y'}) ##. Also show that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial}{\partial y'}(\frac{dF}{dx}) ##.
Relevant Equations
None.
Note that ## \frac{\partial F}{\partial x}=\frac{2x}{2\sqrt{x^2+y'^2}}=\frac{x}{\sqrt{x^2+y'^2}}, \frac{\partial F}{\partial y}=0, \frac{\partial F}{\partial y'}=\frac{2y'}{2\sqrt{x^2+y'^2}}=\frac{y'}{\sqrt{x^2+y'^2}} ##.
Now we have ## \frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}y'+\frac{\partial F}{\partial y'}y"=\frac{x+y'y"}{\sqrt{x^2+y'^2}} ##.
Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{d}{dx}(\frac{y'}{\sqrt{x^2+y'^2}})=\frac{\sqrt{x^2+y'^2}\cdot \frac{d}{dx}(y')-y'\cdot \frac{d}{dx}(\sqrt{x^2+y'^2})}{x^2+y'^2}=\frac{y"\cdot \sqrt{x^2+y'^2}-y'(\frac{x+y'y"}{\sqrt{x^2+y'^2}})}{\sqrt{x^2+y'^2}}=\frac{y"(x^2+y'^2)-y'(x+y'y")}{(x^2+y'^2)^{\frac{3}{2}}} ##.
Also ## \frac{\partial}{\partial y'}(\frac{dF}{dx})=\frac{\partial}{\partial y'}(\frac{x+y'y"}{\sqrt{x^2+y'^2}})=\frac{\sqrt{x^2+y'^2}\cdot \frac{\partial}{\partial y'}(x+y'y")-(x+y'y")\cdot \frac{\partial}{\partial y'}(\sqrt{x^2+y'^2})}{x^2+y'^2}=\frac{\sqrt{x^2+y'^2}\cdot y"-(x+y'y")\cdot (\frac{y'}{\sqrt{x^2+y'^2}})}{x^2+y'^2}=\frac{\sqrt{x^2+y'^2}(\sqrt{x^2+y'^2}\cdot y")-y'(x+y'y")}{\sqrt{x^2+y'^2}}\cdot \frac{1}{x^2+y'^2}=\frac{y"(x^2+y'^2)-y'(x+y'y")}{(x^2+y'^2)^{\frac{3}{2}}} ##.
Therefore, ## \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial}{\partial y'}(\frac{dF}{dx}) ##.
 
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  • #2
As @fresh_42 suggested in another similar question, please try to avoid using y' as a variable, given it often is used to denote the derivative of y. Further, ##\partial F/ \partial y## is confusing , when dealing with ## F(x,y')##.
 
  • #3
WWGD said:
As @fresh_42 suggested in another similar question, please try to avoid using y' as a variable, given it often is used to denote the derivative of y. Further, ##\partial F/ \partial y## is confusing , when dealing with ## F(x,y')##.
I'm guessing that y is a function of a single other variable, say t. If so, y' means ##\frac{dy}{dt}##.
@Math100, please confirm or deny my guess here.
 
  • Like
Likes jim mcnamara and WWGD
  • #4
Mark44 said:
I'm guessing that y is a function of a single other variable, say t. If so, y' means ##\frac{dy}{dt}##.
@Math100, please confirm or deny my guess here.
Yes, I confirm.
 
  • #5
WWGD said:
As @fresh_42 suggested in another similar question, please try to avoid using y' as a variable, given it often is used to denote the derivative of y. Further, ##\partial F/ \partial y## is confusing , when dealing with ## F(x,y')##.
I tried to avoid it too, but the book's problems were all written like that.
 
  • Like
Likes WWGD
  • #6
@Math100, you have ##\frac{\partial F}{\partial y} = 0##. I haven't worked this out, but I don't think this is right.
The other three first partials look OK to me, at a glance. Haven't checked your work on the mixed partials.
 

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