If $u(x(t))$ then what is $\frac{du}{dt}$? My notes say that if $u$ is a function of $x$ and $x$ is a function of $t$ then 
$$\frac{du}{dt}=\frac{∂u}{∂t}+\frac{∂u}{∂x}\frac{dx}{dt}$$
Let's take a simple example where $u=x$ and $x=t$. Then $u=t$ 
So $$\frac{du}{dt}=1$$
$$\frac{du}{dx}=1$$
$$\frac{dx}{dt}=1$$
Which doesn't agree with the above. 
 A: What you have does not quite line up. If we have a function that can be written as a composition $f(t) = u(x(t))$ then the ordinary plain old chain rule gives $$f'(t) = u'(x(t))x'(t),$$ so your example where $u$ and $x$ are both the identity functions gives $$ f'(t) = (1)(1) = 1$$ as it should.
The chain rule you've been given applies when $u$ is a two-variable function of $x$ and $t$ and $x$ is a function of $t.$ This means you can write $f(t) = u(t,x(t))$ and then the rule you've been given can be written $$ f'(t) = \frac{\partial u}{\partial t}(t,x(t))  + \frac{\partial u}{\partial x}(t,x(t))  x'(t).$$ This reduces to the example above when $u$ has no explicit dependence on $t,$ and thus the first term is zero and the $\partial_xu$ in the second term can simply be written as a total derivative $u'$ since it only depends on one variable, $x.$
So your example above should read $$ \frac{\partial u}{\partial t} =0\\\frac{\partial u}{\partial x} =1 \\ \frac{dx}{dt} = 1\\\frac{du}{dt}=1$$ (where, as a commenter noted, it is important to distinguish $\frac{\partial u}{\partial t}$ from $\frac{du}{dt}$) and then the equation checks out. 
