After 2-3 hours thinking about taking derivative of this function:
And finding its answer on the net, don't get how they've taken its derivative using Euler equation. the drivative of this function is this:
This is the Euler equation:
Would you please explain me step by step how they take its drivative.
Added:
This is the whole of that problem and its solution which I'm trying to solve:
Here is the formally correct way to think about the Lagrange equations. First, note that your Lagrangian depends on $x$ only through the dependent "variables" $y$ and $y'$ (I put variables in quotes because $y'$ isn't really what a mathematician would consider a variable). Your Lagrangian
$$f=f(y,y')$$
can be rewritten as a function of two independent variables
$$f=f(v,w):\mathbb{R}^2\to\mathbb{R}$$
$$(v,w)\mapsto v\sqrt{1+w^2}$$
Your problem is asking you to first differentiate $f$ with respect to the second variable,
$$\frac{\partial f}{\partial v}(u,v)$$
then obtain a function of $x$ by plugging in $y(x)$ and $y'(x)$ into the two slots
$$g(x)=\frac{\partial f}{\partial v}(y(x),y'(x))$$
and finally differentiate $g$ with respect to $x$
$$\frac{d g}{dx} = \frac{d}{dx}\Bigg[\frac{\partial f}{\partial v}(y(x),y'(x))\Bigg]$$
Note that I used "$d$" instead of "$\partial$" because $g$ is a function of one variable. Follow this procedure and you'll get the correct answer.
Physicists use incorrect notation sometimes. The notation used in the Lagrange equations is mathematically unsound and confusing for students, although once you get used to it it's quicker than everything I typed out. In your problem, $\frac{\partial ^2}{\partial x \partial z}$ is being used inappropriately. If you evaluated $\frac{\partial ^2}{\partial x \partial z}f$, by equality of mixed partials you could differentiate with respect to $x$ first, and since $f$ is independent of $x$, the whole thing would be zero. On the other hand, applying $\frac{\partial ^2}{\partial x \partial z}$ to $f(x,y(x),v(x))$ makes no sense either, since this is a function of one variable.