Proving Euler-lagrange for single function of single variable with higher derivatives as several functions of one variable One obtains for a functional for single function of single variable with higher derivatives
$   I[f] = \int_{x_0}^{x_1} \mathcal{L}(x, f, f', f'')~\mathrm{d}x ~;~~ 
     f' := \cfrac{\mathrm{d}f}{\mathrm{d}x}, ~f'' := \cfrac{\mathrm{d}^2f}{\mathrm{d}x^2} $
the Euler-Lagrange equation as 
$\cfrac{\partial \mathcal{L}}{\partial f} - \cfrac{\mathrm{d}}{\mathrm{d} x}\left(\cfrac{\partial \mathcal{L}}{\partial f'}\right) + \cfrac{\mathrm{d}^2}{\mathrm{d} x^2}\left(\cfrac{\partial \mathcal{L}}{\partial f''}\right) = 0 $
I want to prove the above equation extending Euler-Lagrange for several functions of one variable i.e.

For a given functional of the form :
$ I[f_1,f_2] = \int_{x_0}^{x_1} \mathcal{L}(x, f_1, f_2,f_1', f_2')~\mathrm{d}x ~;~~ f_i' := \cfrac{\mathrm{d}f_i}{\mathrm{d}x}$
the Euler-Lagrange Equation is
$\frac{\partial \mathcal{L}}{\partial f_i} - \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial \mathcal{L}}{\partial f_i'}\right) = 0 , i=1,2$

This can be achieved by defining $f_1 = f; f_2 =f'$.
But I'm not able to achieve the same equation as for single function of single variable with higher derivatives.
 A: If $f_1$ and $f_2$ are allowed to vary independently, and kept constant at end-point, so $\delta f_1(x_i)=\delta f_2(x_i)=0$, then you obtain the E-L equations stated by using the standard integration by part technique for the linearized variation:
$$ \delta I = \int_{x_0}^{x_1} \left(
\frac{\partial L}{\partial f_1} \delta f_1 + 
\frac{\partial L}{\partial f_2} \delta f_2 + 
\frac{\partial L}{\partial f'_1} \delta f'_1 + 
\frac{\partial L}{\partial f'_2} \delta f'_2 \right) dx=
 \int_{x_0}^{x_1} 
\left((\frac{\partial L}{\partial f_1} - \frac{d}{dx} \frac{\partial L}{\partial f'_1}) \delta f_1 
+(\frac{\partial L}{\partial f_2} - \frac{d}{dx} \frac{\partial L}{\partial f'_2}) \delta f_2 \right) dx
$$
equating to zero for all variations yields the E-L equations.
If here you set $f_2=f_1'$ and impose $\delta f_2= \delta f'_1$ then you may integrate by part once again to get:
 $$ \delta I = \int_{x_0}^{x_1} \left(\frac{\partial L}{\partial f_1} - \frac{d}{dx} \frac{\partial L}{\partial f'_1} 
-\frac{d}{dx} \frac{\partial L}{\partial f_2} + \frac{d^2}{dx^2} \frac{\partial L}{\partial f'_2} \right) \delta f_1  dx
$$
hereby obtaining the second order formulation by equating to zero. 
