My professor was deriving the Strong Euler-Lagrange equation today and I was unsure about the following statement (no calculus of variations knowledge required)
$$g(h) = I[y(x)+hu(x)] := \int_{x_1}^{x_2}f\big(y(x)+hu(x),y'(x)+hu'(x),\ldots,y^{(n)}(x)+hu^{(n)}(x)\big)dx$$
$$g'(0) = \int_{x_1}^{x_2}\frac{\partial f}{\partial y}u+\frac{\partial f}{\partial y'}u'+\ldots+\frac{\partial f}{\partial y^{(n)}}u^{(u)}dx$$
I tried computing this from scratch by defining $w_i(x) = y^{(i)}(x) + hu^{(i)}(x), 0 \leq i \leq n$ giving
$$\int_{x_1}^{x_2}\frac{d}{dh}f(w_0,w_1,\ldots,w_n)dx = \int_{x_1}^{x_2} \sum_{i=0}^n \frac{\partial f}{\partial w_i}\frac{\partial w_i}{\partial h}dx = \int_{x_1}^{x_2} \sum_{i=0}^n \frac{\partial f}{\partial w_i}u^{(i)}dx$$
I am under the impression that each of these partial derivatives are $\displaystyle \frac{\partial f}{\partial w_i} = \frac{\partial f}{\partial w_i}(w_0,\ldots,w_n)$ and evaluation at $h=0$ gives $w_n = y^{(n)}$, resulting in $\displaystyle \frac{\partial f}{\partial w_i}(y,\ldots,y^{(n)})$ which is nice and as expected, but why does this affect what we are taking the partial derivative of $f$ w.r.t with?
Where does $\displaystyle\frac{\partial f}{\partial y^{(i)}}$ come in from?
If I compute for $f(x) = x^2$ the derivative $\frac{df}{dx}$ and evaluate at $x=0$ I certaintly do not write $\frac{df}{d0} = 0$.
What is the error I am making? Apologies if this is a really silly and trivial misunderstanding.
With $w_i=v^{(i)}+hu^{(i)}$ and some $g(x,y)$
$$\left.\frac{\partial g(w_i,y)}{\partial w_i}\right|_{h=0}=\frac{\partial g(v^{(i)},y)}{\partial v^{(i)}}$$
Added
$g(x)=x^2$
$g(w_i)=w_i^2=(v^{(i)}+hu^{(i)})^2$
$g(v^{(i)})=(v^{(i)})^2$
$\dfrac{\mathbb dg(w_i)}{\mathbb dw_i}=2w_i=2(v^{(i)}+hu^{(i)})$
$\left.\dfrac{\mathbb dg(w_i)}{\mathbb dw_i}\right|_{h=0}=2v^{(i)}=\dfrac{\mathbb dg(v^{(i)})}{\mathbb dv^{(i)}}$