How to find ∂u/∂x and ∂u/∂t of this?

Question

We are given $$u(x, t) = f(x − t*u(x, t))$$

How do we determine $$\frac{\partial u}{\partial x} \quad\text{and}\quad \frac{\partial u}{\partial t}? $$

Attempt:

$$\frac{\partial u}{\partial x} = f'(x-t*u(x,t))$$

$$\frac{\partial u}{\partial t} = -u(x,t)*f'(x-t*u(x,t)) $$

Help appreciated, thanks!

Answer 1

We have $ u(x,t)=f(z) $ with $z=x-tu$. Using the chain rule and supposing that $x, t$ are independent variables ( so that $\frac{\partial t}{\partial x}=0$), we have:

$$ \frac{\partial u}{\partial x}= f'(z)\frac{\partial z}{\partial x}=f'(z)\frac{\partial (x-tu)}{\partial x}= $$ $$ =f'(z)\left(1-\frac{\partial t}{\partial x}u-t\frac{\partial u}{\partial x}\right)=f'(z)\left(1-t\frac{\partial u}{\partial x}\right) $$ that gives $$ \frac{\partial u}{\partial x}\left[1+tf'(z) \right]=f'(z) $$

so we have: $$ \frac{\partial u}{\partial x}=\frac{f'(z)}{\left[1+tf'(z) \right]}=\frac{f'(x-tu)}{\left[1+tf'(x-tu) \right]} $$

You can do the same for the derivative with respect to $t$

Answer 2

given $$u(x,t)=f(x-tu(x,t))$$ then we get by the chain rule $$\frac{\partial u(x,t)}{\partial x}=f_x(x-tu(x,t))(1-tu_x(x,t))$$ and $$\frac{\partial u(x,t)}{\partial t}=f_t(x-tu(x,t))(-u(x,t)-tu_t(x,t))$$

Answer 3

By the chain rule,

$$\dfrac{\partial u}{\partial x}=f'(x-tu(x,t))\mathbf{\dfrac{\partial}{\partial x}\left(x-tu(x,t)\right)};$$ Can you conclude and to the same for $\partial u/\partial t$?