# Why is this is the derivative?

We are using the Euler equation to calculate the minimum:

Euler equation: $-\frac{d}{dt}\hat{L}_{x'}(t) + \hat{L}_x(t) = 0$

We have the following $L = 12tx + x'^2$ ($x$ is a function of $t$)

Now calculating these derivatives my book says it equals $-2x'' + 12t = 0$

Can anybody please explain how they came to this answer?

Hints. Rewrite as follows: $$-\frac{d}{dt} \frac{\partial L}{\partial x'} + \frac{\partial L}{\partial x} = 0 \qquad ; \qquad L = 12tx + (x')^2$$ What is $\frac{\partial L}{\partial x'}$ , what is $-\frac{d}{dt} \frac{\partial L}{\partial x'}$ then, and, at last, what is $\frac{\partial L}{\partial x}$ ?
We first differentiate with respect to $t$, that is, $x''(t)$, then we differentiate with respect to $x'$, that is $1$, using the chain rule we get $\frac{d}{dt}\frac{d}{dx}\hat{L} = 1\cdot x'' = x''$