Help taking a partial derivative

[docnet]

Homework Statement

I need to compute this partial derivative

Relevant Equations

$\partial_tu\big[(t,x-t\kappa V)\big]$

Hi all, I was wondering is if the following partial derivative can be computed without a specific $u(t,x)$

$$\partial_tu\big[(t,x-t\kappa V)\big]$$

I was thinking it can't be done, because we could have

$$u_a(t,x)=tx \Rightarrow \partial_tu\big[(t,x-t\kappa V)\big]=\partial_t\big[tx-t^2\kappa V\big]=x-2t\kappa V$$
$$u_b(t,x)=t+x \Rightarrow \partial_tu\big[(t,x-t\kappa V)\big]=\partial_t\big[t+x-t\kappa V\big]=1-\kappa V$$

so there is no universal formula for $\partial_tu\big[(t,x-t\kappa V)\big]$, which depends on the function $u(t,x)$.

 

[PeroK]

You mean that the partial derivative of a function depends on the function? Is that a surprise?

 

[docnet]

You mean that the partial derivative of a function depends on the function? Is that a surprise?

Thank you. If we have a transformation of $u(t,x)$ defined by $u_a(t,x)\Rightarrow u(2t,x)$ then we can make a statement like
$$\partial_t[u_a(t,x)]=\partial_tu(2t,x)\cdot \partial_t(2t)$$$$=2\cdot \partial_tu(2t,x)=2\partial_tu_a(t,x)$$
The partial derivative of the transformed function scales by a multiple of 2.

I needed some thinking out loud to convince myself that there isn't a similar result for functions like $u(t,x-t)$

 

[docnet]

okay, i think i see what is confusing me. i have been writing it wrong.

could you tell me if the following statement is correct, and the statement in post is #3 wrong?

$$\partial_tu_a(t,x)=\partial_tu(2t,x)=\partial_t u(t,x) \cdot \partial_t(2t)$$
$$=2 \partial_tu(t,x)$$

 

[PeroK]

okay, i think i see what is confusing me. i have been writing it wrong.

could you tell me if the following statement correct, and the statement in post #3 wrong?

$$\partial_tu_a(t,x)=\partial_tu(2t,x)=\partial_t u(t,x) \cdot \partial_t(2t)$$
$$=2 \partial_tu(t,x)$$

That's not right. The general form of the chain rule is: $$h(x) = f(g(x)) \ \Rightarrow \ h'(x) = f'(g(x))g'(x)$$ Note that the derivative of $f$ is evaluated at $g(x)$. This generalises to the multi-variable case: $$\partial_t u(f(t), x) = \partial_t u(f(t), x) f'(t)$$ Now, if you have a function of the form $u_a(t, x) = u(f(t), g(t, x))$, then you need to be careful about your notation and what things means. In this case, we have:
$$\partial_t u_a(t, x) = \partial_t u(f(t), g(t, x))f'(t) + \partial_x u(f(t), g(t, x))\partial_tg(t, x)$$
For more detail, see my Insight on the multi-variable chain rule:

https://www.physicsforums.com/insights/demystifying-chain-rule-calculus/

 

[docnet]

That's not right. The general form of the chain rule is: $$h(x) = f(g(x)) \ \Rightarrow \ h'(x) = f'(g(x))g'(x)$$ Note that the derivative of $f$ is evaluated at $g(x)$. This generalises to the multi-variable case: $$\partial_t u(f(t), x) = \partial_t u(f(t), x) f'(t)$$ Now, if you have a function of the form $u_a(t, x) = u(f(t), g(t, x))$, then you need to be careful about your notation and what things means. In this case, we have:
$$\partial_t u_a(t, x) = \partial_t u(f(t), g(t, x))f'(t) + \partial_x u(f(t), g(t, x))\partial_tg(t, x)$$
For more detail, see my Insight on the multi-variable chain rule:

https://www.physicsforums.com/insights/demystifying-chain-rule-calculus/

awesome write up! thank you.