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.

• Mabye if I write your expression as $$\dfrac{d}{dx} = \dfrac{\partial}{\partial x} +\dfrac{dy}{dx}\dfrac{\partial}{\partial y} + \dfrac{d y'}{dx}\dfrac{\partial}{\partial y'}$$ It would help. Aug 18, 2016 at 8:50
• @Chinny84 I appreciate your reply. How can d/dx alone be that complicated? Isn't d/dx simply just differentiating with respect to x? I'm guessing it has something to do with y being a function of x but...not sure. Am I missing something here? Sorry if it feels like a trivial question,but I really do not know. Aug 18, 2016 at 8:55

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.