# Derivative of Functional

#### danong

hi there, i am actually studying about functional equation.
I got stucked with some derivatives problem,
and where i could find nowhere to refer or study from,
because it seems it is out of university book level.

my question is this :

what does it means by taking derivative with respect to partial derivative?
can anyone visualize this idea to me?
because i couldn't figure out the term with respect to partial derivative,
when it comes to a functional equation,
of which is a differential equations.

For e.g : F(x,y(x),y'(x),y''(x)) , find $$\displaystyle \frac{d}{dx}$$ of $$\displaystyle \partial$$ F(x,y(x),y'(x),y''(x)) / $$\displaystyle \partial$$ y' .

How can we write the full solution with partial derivative respect to y' ? and how bout y''?

#### Drexel28

MHF Hall of Honor
hi there, i am actually studying about functional equation.
I got stucked with some derivatives problem,
and where i could find nowhere to refer or study from,
because it seems it is out of university book level.

my question is this :

what does it means by taking derivative with respect to partial derivative?
can anyone visualize this idea to me?
because i couldn't figure out the term with respect to partial derivative,
when it comes to a functional equation,
of which is a differential equations.

For e.g : F(x,y(x),y'(x),y''(x)) , find $$\displaystyle \frac{d}{dx}$$ of $$\displaystyle \partial$$ F(x,y(x),y'(x),y''(x)) / $$\displaystyle \partial$$ y' .

How can we write the full solution with partial derivative respect to y' ? and how bout y''?
I assume you're talking about a standard Calculus of Variations problem with a functional of the form $$\displaystyle J=\int_a^b F(x,y,y',y'')dx$$. Now, for this specific case the variational derivative is of the form $$\displaystyle F_y-\frac{\partial }{\partial x}F_{y'}=0$$. Now, I agree this can be confusing at first but what that really means is call $$\displaystyle F(x,y,y',y'')=F(t,u,v,w)$$. So, for example if $$\displaystyle F(x,y,y',y'')=x^2y+y'+\frac{y'}{y''}$$ you would get $$\displaystyle F(t,u,v,w)=t^2u+v+\frac{v}{w}$$. Now, the Euler-Lagrange equation then says take $$\displaystyle F_u$$ so in this case $$\displaystyle F_u=t^2$$ and $$\displaystyle F_v=1+\frac{1}{v}$$. But! You have to sub back in what that really means, i.e. $$\displaystyle F_y=x^2,F_{y'}=1+\frac{1}{y''}$$. Now, treating these again as actual functions of $$\displaystyle x$$ you compute $$\displaystyle F_y-\frac{d}{dx}F_{y'}$$. The point is that $$\displaystyle F_y,F_{y'}$$ intend you to treat $$\displaystyle F$$ as a function of the independent variables $$\displaystyle x,y,y',y''$$ and differentiate with respect to the specified one. But! Once done with that the $$\displaystyle \frac{d}{dx}$$ intends you to treat $$\displaystyle F_y,F_{y'}$$ again as a function of functions of $$\displaystyle x$$.

Write back if you need more clatification.

#### Jester

MHF Helper
hi there, i am actually studying about functional equation.
I got stucked with some derivatives problem,
and where i could find nowhere to refer or study from,
because it seems it is out of university book level.

my question is this :

what does it means by taking derivative with respect to partial derivative?
can anyone visualize this idea to me?
because i couldn't figure out the term with respect to partial derivative,
when it comes to a functional equation,
of which is a differential equations.

For e.g : F(x,y(x),y'(x),y''(x)) , find $$\displaystyle \frac{d}{dx}$$ of $$\displaystyle \partial$$ F(x,y(x),y'(x),y''(x)) / $$\displaystyle \partial$$ y' .

How can we write the full solution with partial derivative respect to y' ? and how bout y''?
It might be best to look at an example. Consider

$$\displaystyle L(x,y,y') = \frac{\sqrt{1 + y'^2}}{\sqrt{y}}$$.

Then

$$\displaystyle \frac{\partial L}{\partial y'} = \frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2}}$$ (treat $$\displaystyle x,y$$ and $$\displaystyle y'$$ as all independent variables). Then

$$\displaystyle \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = \frac{d}{dx} \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2} } \right)$$

$$\displaystyle = \frac{\partial}{\partial x} \underbrace{ \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2 }}\right)}_{\text{this will} = 0\, \text{- no}\,x }$$ $$\displaystyle + \frac{\partial}{\partial y} \underbrace{ \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2 }}\right)}_{\text{hold}\, y'\, \text{fixed} } y'$$ $$\displaystyle + \frac{\partial}{\partial y'} \underbrace{ \left( \frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2}} \right)}_{\text{hold}\, y\, \text{fixed} } y''$$.

For the last two derivatives, treat $$\displaystyle y$$ and $$\displaystyle y'$$ and independent.

(A little slow)