Derivative with respect to $y'$ in the Euler-Lagrange differential equation I am having trouble understanding the $ \dfrac{\partial L}{\partial y'} $ part in Euler-Lagrange Equation.
For example, if $ L = y^2(z) $, what is the symbolic expression for 
$ \displaystyle\frac{\partial (y^2(z))}{\partial (\frac{\partial y}{\partial z})} $?
 A: To expand Rasmus' answer, a lagrangian can be seen as a function $\mathcal{L}=\mathcal{L}(x,y,p)$ of three independent variables. Then you will mainly work with $p=y'$, but the notation
$$
\frac{\partial\mathcal{L}}{\partial y'}
$$
should be understood as a shorthand for
$$
\frac{\partial \mathcal{L}}{\partial p}.
$$
More concretely, if (say) $\mathcal{L}(x,y,y')=\frac{1}{2}|y'|^2 + \frac{1}{2}y^2$, then
$$
\frac{\partial\mathcal{L}}{\partial y'} = y',
$$
since
$$
\frac{\partial}{\partial p} \left( \frac{1}{2}|p|^2 + \frac{1}{2}y^2 \right) = p.
$$
A: That would simply be $0$ because $y'$ is regarded as a variable independent of $y$.
A: When extremizing
$$I\left[y,\dot{y}\right]=\int\mathcal{L}(y,\dot{y},t)dt$$
One finds the difference between two near values and declares that the functional has been minimized when there is no change to first order. (assume fixed end points)
$$\Delta I\left[y,\dot{y}\right]=I\left[y+\delta y,\dot{y}+\delta \dot{y}\right]-I\left[y,\dot{y}\right]$$
$$\Delta I\left[y,\dot{y}\right]=\int(\mathcal{L}(y+\delta y,\dot{y}+\delta\dot{y},t)-\mathcal{L}(y,\dot{y},t))dt$$
Taking a taylor expansion of the integrand to first order, (note we will need to differentiate with respect to $\dot{y}$.
$$\delta I\left[y,\dot{y}\right]=\int(\delta y\frac{\partial \mathcal{L}}{\partial y}+\delta\dot{y}\frac{\partial \mathcal{L}}{\partial \dot{y}})dt$$
Integrating by parts (with vanishing boundary term)
$$\delta I\left[y,\dot{y}\right]=\int\delta y(\frac{\partial \mathcal{L}}{\partial y}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{y}})dt$$
Now if $\delta I=0$ for all admissible variations $\delta y$ then we have that the integrand must vanish identically
$$\frac{\partial \mathcal{L}}{\partial y}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{y}}=0$$
