Partial derivatives - how do I get rid of u from this expression?

Let $u(x,t)=\phi(x+ct)+\psi(x-ct)$ where $\phi$ and $\psi$ are arbitrary functions. Find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial t}$ in terms of the partial derivatives of $\phi$ and $\psi$.

I've written that $u=u(\phi,\psi)$ and $\phi=\phi(x,t), \psi=\psi(x,t)$ and using the chain rule I ended up with...

$\frac{\partial u}{\partial x}_t$=$\frac{\partial u}{\partial \phi}_\psi \frac{\partial \phi}{\partial x}_t + \frac{\partial u}{\partial \psi}_\phi \frac{\partial \psi}{\partial x}_t$
$\frac{\partial u}{\partial t}_x$=$\frac{\partial u}{\partial \phi}_\psi \frac{\partial \phi}{\partial t}_x + \frac{\partial u}{\partial \psi}_\phi \frac{\partial \psi}{\partial t}_x$

Is that it? The question asks for an answer in terms of partial derivatives of $\phi$ and $\psi$ but I still have partial derivatives of $u$... I don't know how to get rid of that though since I don't know what $\phi$ and $\psi$ are. And my answer has nothing to do with the fact that $\phi$ and $\psi$ are functions of $x+ct$ and $x-ct$ so I'm getting the feeling I'm missing something... but I'm not sure what. Could someone give me a hint? Thanks!

∂u/∂x = φ'(x+ct) + ψ'(x-ct)
∂u/∂t= c φ'(x+ct) - c ψ'(x-ct)

PS this is the general equation of a one dimensional wave, transversing with speed c

Oh wow, I made a right mess of that then. Thank you! Rest of the question makes a lot more sense now too...

My pleasure as always

What course are you doing and what place if you do not mind asking.

"learn by doing"

Earth Sciences, Oxford. Our lecturer has gone through partial fractions extremely quickly (particularly bearing that most of the cohort hasn't done Further Maths) so we've very much had to "learn by doing" in the problem sheets - which I've actually found quite interesting, but there were a couple of questions I was stuck on and I'd like to work out how to do them before my tutorial.

∂u/∂x = φ'(x+ct) + ψ'(x-ct)
∂u/∂t= c φ'(x+ct) - c ψ'(x-ct)

PS this is the general equation of a one dimensional wave, transversing with speed c

Not wishing to muddy the waters, but I'd appreciate your thoughts on this.

What is "φ'(x+ct)" here?

My thinking would be $\dfrac{\partial \phi}{\partial (x+ct)}$, which is rather hidden using the apostrophe symbolism.

Is that correct?

Not wishing to muddy the waters, but I'd appreciate your thoughts on this.

What is "φ'(x+ct)" here?

My thinking would be $\dfrac{\partial \phi}{\partial (x+ct)}$, which is rather hidden using the apostrophe symbolism.

Is that correct?

correct

Yes, my lecturer says that the prime denotes differentiation w.r.t the variable inside the function

...

A subsequent thought: We could write this as $\dfrac{d \phi}{d (x+ct)}$ then, since "x+ct" is a single variable.

Yes?

imagine the phi function is x^3 or u^3 ot v^3 ....

then (x+ct)^3 will differentiate wrt t

will differentiate to 3(x-ct)^2 times c

'Fraid you've lost me there. I'm not even clear whether you're agreeing or disagreeing with me.

'Fraid you've lost me there. I'm not even clear whether you're agreeing or disagreeing with me.