Basic of Partial Differential Equation I pretty new to calculus and I am trying to understand the following transformations:


*

*$2uu_{t} = \frac{\partial }{\partial t}u^{2} $

*$2u_{t}u_{tt} = \frac{\partial }{\partial t}u_{t}^{2} $

*$2uu_{xx} = 2\frac{\partial }{\partial x}\left ( uu_{x} \right ) -2u_{x}^{2} $


The first two, I believe that are related to the power function rule $\frac{\partial }{\partial x}u^{n} = (n)u^{n-1}u_{x}$. 
But the last item, I'm not sure... someone can help me understand how I get this result?
 A: Correct for the first two. For the third you need the product rule, which says that $\frac{\partial}{\partial x}(uv)=u_xv+uv_x$ if $u,v$ are functions of $x$. Se we have:
$$2\frac{\partial}{\partial x}(uu_x)-2u_x^2$$
$$=2(u_x\cdot u_x+u\cdot u_{xx})-2u_x^2$$
$$=2uu_{xx}$$
A: Just use the product rule on the first term on the right hand side: $2\frac{\partial}{\partial x}(u u_x) = 2u_x u_x+2uu_{xx} = 2u_x^2+2uu_{xx}$.
A: If you start from the left hand side, you can integrate it and use by-part:
$$\int 2uu_{xx}dx=2uu_x-2\int u^2_xdx$$
Then differentiate both sides with respect to x and get:
$$2uu_{xx}=2\frac{\partial}{\partial x}(uu_x)-2u^2_x$$
as desired.  

If you start from the right hand side, you can use the product rule:
$$2\frac{\partial}{\partial x}(uu_x)-2u^2_x$$
$$=(2u_xu_x+2uu_{xx})-2u^2_x$$
$$=2uu_{xx}$$
as desired.
A: Let us prove that 
$2uu_{xx} = 2\frac{\partial }{\partial x}\big ( uu_{x} \big ) -2u_{x}^{2}$. 
First, we elaborate the first term of the right hand side :
$$
2\,\frac{\partial }{\partial x}\big ( uu_{x} \big ) = 
2 \big( u_x \cdot u_x + u \cdot u_{xx}\big) = 
\boxed{\ 2\left( u_x^2 + u  u_{xx}\right) \ }
$$
Second, we subtract $2u_{x}^{2}$ from the result:
$$
2\frac{\partial }{\partial x}\Big ( uu_{x} \Big ) -2u_{x}^{2} = 
 2\left( u_x^2 + u  u_{xx}\right) -2u_{x}^{2} = 2 u  u_{xx}.
$$
Thus we proved that $2uu_{xx} = 2\frac{\partial }{\partial x}\big ( uu_{x} \big ) -2u_{x}^{2}$.
