Can anyone please verify/confirm these derivatives?

[Math100]

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, $\frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial}{\partial y'}(\frac{dF}{dx})$.

 

[WWGD]

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 $F(x,y')$.

 

[Mark44]

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.

 

[Math100]

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.

 

[Math100]

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.

 

[Mark44]

@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.