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

[Screwdriver]

Homework Statement

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

[tex]\frac{\partial u}{\partial x}[/tex]

In terms of [tex]f[/tex] and its derivatives.

Homework Equations

PD Chain rule.

The Attempt at a Solution

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

[tex]\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}u(x(f),y(f))\frac{dx}{d f}(x(f))[/tex]

[tex]\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)[/tex]

[tex]\frac{\partial u}{\partial x}=f'(x+ct)[/tex]

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

[tex]\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)[/tex]

?

 

[Dick]

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

[tex]\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)[/tex]

?

Yes, exactly.

 

[Screwdriver]

Okay. Thanks again