Partial Derivative Chain Rule for u(x, t) in terms of f and its Derivatives

[Screwdriver]

Homework Statement

Say that f(x) is some function whose second derivative exists and say u(x, t)=f(x + ct) for c > 0. Determine

\frac{\partial u}{\partial x}

In terms of f and its derivatives.

Homework Equations

PD Chain rule.

The Attempt at a Solution

Say that x and y are both functions of f, ie. x = x(f) and y = y(f). Then,

\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}u(x(f),y(f))\frac{dx}{d f}(x(f))

\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}u(x,y)\frac{dx}{d f}(x(f))=\frac{\partial }{\partial x}f(x+ct)\frac{dx}{d f}(x+ct)

\frac{\partial u}{\partial x}=f'(x+ct)

Does that make any sense? It's mainly the notation that I don't understand, especially the "let x and y be functions of the same variable part."

 

[Dick]

No. That doesn't make any sense at all. You said f is a function of x+ct. That x would then be a function of f is crazy talk. What you've got there is just finding the partial derivative of f(x+ct) with respect to x where x and t are the independent variables.

 

[Screwdriver]

Thanks for the reply. So should I just disregard that extra function part and just write

\frac{\partial }{\partial x}u(x,t)=\frac{\partial }{\partial x}f(x+ct)=\frac{\partial }{\partial x}f(x+ct)\frac{d }{d x}(x+ct)=f'(x+ct)

?

 

[Dick]

Thanks for the reply. So should I just disregard that extra function part and just write

\frac{\partial }{\partial x}u(x,t)=\frac{\partial }{\partial x}f(x+ct)=\frac{\partial }{\partial x}f(x+ct)\frac{d }{d x}(x+ct)=f'(x+ct)

?

Yes, exactly.

 

[Screwdriver]

Okay. Thanks again