How to differentiate with respect to a derivative

[Jozefina Gramatikova]

Hi guys, I am reading my lecture notes for Mechanics and Variations and I am trying to understand the maths here. From what I can see there we differentiated with respect to a derivative. Could you tell me how do we do that? Thanks

 

[mfb]

Treat it like a regular variable. You can introduce z=y' if that helps.

 

[Jozefina Gramatikova]

how do we differentiate y with respect to y' then?

 

[fresh_42]

how do we differentiate y with respect to y' then?

Substitute $y'=z$, differentiate along $\partial z$, and re-substitute $z=y'$.

 

[Ray Vickson]

how do we differentiate y with respect to y' then?

You just have two variables, which I am going to call $a$ and $b$ (instead of $y$ and $y'$). Thus, you have
$$\tau(a,b) = \frac{\sqrt{1+b^2}}{\sqrt{-2 g a}}$$ Now the derivatives $\partial \tau(a,b)/ \partial a$ and $\partial \tau(a,b) / \partial b$ are just perfectly ordinary partial derivatives.

 

[Delta2]

how do we differentiate y with respect to y' then?

There is no generic formula if that's what you wonder about

I mean for example if $y(x)=x^2$ then $y'(x)=2x$ and hence $y(y')=(y')^2/4$ hence $\frac{\partial y}{\partial y'}=\frac{2}{4}y'$
On the other hand if $y(x)=\sin x$ then $y'(x)=\cos x$ and hence $y(y')=\sqrt{1-y'^2}$ hence $\frac{\partial y}{\partial y'}=\frac{-2y'}{2\sqrt{1-y'^2}}$

 

[PeroK]

how do we differentiate y with respect to y' then?

The fundamental idea behind the calculus of variations is to study the integrand as an abstract function of the variables involved. In this case $y$ and $y'$, leaving to one side that as physical variables they are related.

As others have said, therefore, you treat $y$ and $y'$ as independent variables in order to generate the E-L equations.

 

[Jozefina Gramatikova]

Ok, thank you very much, guys. I got it :)