Can someone explain me how using the implicit function theorem get the following result
Given $u_0(x) = u(x,0)$, $f'(u)$ and the implicit function $$F = u(x,t) - u_0(x-f'(u)t) = 0$$
compute the derivative of $u$ w.r.t. $x$
$$\frac{\partial u}{\partial x} = \frac{-u'_0}{1+u'_0f''(u)t}$$
$$u(x,t)-u_0(x-f'(u)t)=0\Rightarrow u=u_0(x-f'(u)t)\Rightarrow\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}u_0(x-f'(u)t)$$ The right side is equal to (using chain rule): $$\frac{\partial}{\partial x}u_0(x-f'(u)t)=u'_0\cdot(1-f''(u)t\frac{\partial u}{\partial x})$$ So we have $$\frac{\partial u}{\partial x}=u'_0\cdot(1-f''(u)t\frac{\partial u}{\partial x})$$ Rearranging it yields $$\frac{\partial u}{\partial x}=\frac{u'_0}{1+u'_0f''(u)t}$$