Euler Lagrange Equation differentiate x I want to find what this is:
$$ \dfrac{d}{dx}\left(g(x) \dfrac{y'}{\sqrt{1+y'^2}}\right) $$
I've attempted it and got:
$$ = \dfrac{g(x)}{\sqrt{1+y'^2}^3} - \dfrac{g'(x)}{\sqrt{1+y'^2}}$$
but seems like i'm getting the wrong answer to what i'm supposed to get.
I've seen the question Euler-Lagrange equation: Differentiation with respect to x, and I still don't understand why $$\dfrac{d}{dx} = \dfrac{\partial}{\partial x} + y'\dfrac{\partial}{\partial y} + y''\dfrac{\partial}{\partial y'}$$
How 'chain rule' work here ??? I don't understand it as I have been off maths for a while, and am struggling when notations get more complex. Thanks.
 A: As the user who provided the second answer to the question you cited explained, the function $f = f(x, y, y')$ where $y = y(x)$. Then the total derivative of $f$ with respect to $x$ is $$\dfrac{d}{dx}f(x, y, y') = \dfrac{\partial f}{\partial x} +  \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial x} +  \dfrac{\partial f}{\partial y'} \dfrac{\partial y'}{\partial x}.$$ Because $ \dfrac{\partial f}{\partial x}$,  $\dfrac{\partial f}{\partial y}$, and $ \dfrac{\partial f}{\partial y'}$ are also functions of $x$, $y$, and $y'$, their total derivatives with respect to $x$ take the same form. So in general (for the specific question in the link you gave), $$\dfrac{d}{dx} = \dfrac{\partial}{\partial x} +  \dfrac{\partial}{\partial y} \dfrac{\partial y}{\partial x} +  \dfrac{\partial}{\partial y'} \dfrac{\partial y'}{\partial x}.$$ Now for your question, $\dfrac{d}{dx}$ is slightly different. Let $$f = g(x)\dfrac{y'}{\sqrt{1 + y'^{2}}}.$$ What is $f$ a function of? Try to use the same idea as above to write what $\dfrac{d}{dx}$ is for your problem. Hope that helps.
