Scaling solution of the diffusion equation I am studying the book "A guide to first-passage processes", by Sidney Redner, and at page 14 he derives a solution of the diffusion equation
$\dfrac{\partial c(x,t)}{\partial t} = D \dfrac{\partial^2c(x,t)}{\partial x^2} \,,$
based on a scaling ansatz of the form
$c(x,t) = \dfrac{1}{X(t)} f[x/X(t)]  \,.$
The book says: "substituting this ansatz into the diffusion equation gives
$X(t) \dot{X}(t) = -D \dfrac{f''(u)}{f(u) + uf'(u)} \,,$ $\quad$ where $u=\dfrac{x}{X(t)}$." 
The prime denotes differentiation w.r.t. $u$ and the overdot denotes the time derivative. Can anybody please give me a hand on how to arrive there?
EDIT: As pointed out in the comments and in the answers, this is simply repeated applications of the chain rule. What is important to keep in mind is that
$\dfrac{\partial f(u)}{\partial t} = \dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial t} = f'\dfrac{\partial u}{\partial t}\,,$
$\dfrac{\partial f(u)}{\partial x} = \dfrac{\partial f}{\partial 
u}\dfrac{\partial u}{\partial x}= f'\dfrac{\partial u}{\partial x}\,,$
and then everything works out smoothly. Thank you everybody.
 A: $$c(x,t) = \dfrac{1}{X(t)} f[x/X(t)]  \,.$$
Differentiate twice wrt $x$:
$$\dfrac {dc(x,t)}{dx} = \dfrac{1}{X^2(t)} f_u[u]  \,.$$
$$\dfrac {d^2c(x,t)}{dx^2} = \dfrac{1}{X^3(t)} f_{uu}[u]  \,.$$
Differentiate wrt $t$:
$$\dfrac {dc(x,t)}{dt} = \dfrac{1}{X^2(t)} \left ( -f_u[u] \dfrac {xX_t(t)} {X(t)}- X_t(t)f[u] \right )$$
$$\dfrac {dc(x,t)}{dt} = \left ( -f_u[u] \dfrac {xX_t(t)} {X^3(t)}- \dfrac{X_t(t)f[u] }{X^2(t)} \right )$$
The diffusion equation is:
$$\dfrac{\partial c(x,t)}{\partial t} = D \dfrac{\partial^2c(x,t)}{\partial x^2} \,,$$
$$f_u[u] \dfrac {xX_t(t)} {X^3(t)}+ \dfrac{X_t(t)f[u] }{X^2(t)} =-\dfrac{D}{X^3(t)} f_{uu}[u]  \,.$$
$$f_u[u]  {xX_t(t)} + {X_t(t)f[u] }{X(t)} =-{D}f_{uu}[u]  \,.$$
The result follows:
$$X_t(t){X(t)}\left (f_u[u] u + f[u]  \right )=-{D}f_{uu}[u]  $$
$$ \boxed {X_t(t){X(t)}=-{D} \dfrac  {f_{uu}[u] }{\left (f_u[u] u + f[u] \right )}}$$I think that the derivative at the denominator is taken according to the variable $u$.  At the numerator too, its taken with respect to u. Where:
$$u=\frac x {X(t)}$$
