Chain Rule in Calculus of Variations

Question

When you take the derivative of the function $F(y+a\eta, y'+a\eta, x)$ with respect to $a$, you first take the derivative of $F$ with respect to $y+a\eta$ then multiply it with the derivative of $y+a\eta$ with respect to $a$, right? But how do you take the derivative of $F$ with respect to $y+a\eta$? We do not know the function exactly. In the book I am studying (Taylor Classical Mechanics), the result is given as η.f'(y)+η.f'(y')

Answer 1

If we make the substitutions

$\xi=y+a\eta$

$\zeta=y'+a\eta$

Then

$$F(y+a\eta,y'+a\eta,x)=F(\xi,\zeta,x)$$

If we would like to find the derivative of $F$ wrt $a$ and assuming $x$ does not vary with $a$

$$\frac{\partial F}{\partial a}=\frac{\partial F}{\partial\xi}\frac{\partial\xi}{\partial a}+\frac{\partial F}{\partial \zeta}\frac{\partial\zeta}{\partial a}$$

Without the function, there's no telling what the derivative is. The most you could write is

$$\frac{\partial F}{\partial(y+a\eta)}=\frac{\partial F}{\partial\xi}$$

Meaning differentiate wrt $y+a\eta$ as if it were a single variable

Answer 2

On the previous page in Chapter 6.2, Taylor mentions "we must just check that the derivative dS/d$\alpha$ is zero when $\alpha=0$," where $S=\int_{x_1}^{x_2}f\left[y\left(x\right),\ y'\left(x\right),\ x\right]dx$. So I believe that in addition to applying the chain rule, Taylor evaluates $\frac{\partial f\left(y+\alpha \eta,\ y'+\alpha \eta',\ x\right)}{\partial \alpha}$ at $\alpha=0$. Then $Y(x)=y(x)+\alpha \eta(x)$ gives $$\frac{\partial f\left(y+\alpha \eta,\ y'+\alpha \eta',\ x\right)}{\partial \alpha}\Big|_{\alpha=0}=\frac{\partial Y}{\partial \alpha} \frac{\partial f}{\partial Y}\Big|_{\alpha=0}+\frac{\partial Y'}{\partial \alpha} \frac{\partial f}{\partial Y'}\Big|_{\alpha=0}=\eta \frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}.$$