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.
$$c(x,t) = \dfrac{1}{X(t)} f[x/X(t)] \,.$$ 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: $$u=\frac x {X(t)}$$