Given two twice-differentiable functions $f$ and $g$, use the chain rule to express the following derivatives of $u(x,t)=f(x+4t)+g(x-4t)$. In terms of

Question

Given two twice-differentiable functions $f$ and $g$, use the chain rule to express the following derivatives of \begin{align*} u(x,t)=f(x+4t)+g(x-4t). \end{align*} In terms of $f, g, f', g', f'', g''$.

a) $u_x$ y $u_t$

b) $u_{xx}$ y $u_{tt}$

c) $u_t+4u_x$

d) $u_{tt}-16u_{xx}$

I'm a little confused with the notation, I guess. For example

$u_x=\frac{\partial }{\partial x}\left(f(x+4t)\right)+\frac{\partial }{\partial x}\left(g(x-4t)\right)$

But how do I calculate $\frac{\partial }{\partial x}\left(f(x+4t)\right)$ and $\frac{\partial }{\partial x}\left(g(x-4t)\right)$

They can explain. Thank you.

I think it is:

$u_x(x, t)=f'(x+4t)+g'(x-4t)$

$u_t(x, t)=4f'(x+4t)-4g'(x-4t)$

Answer 1

a) We have \begin{align*} u_x=f'(x+4t)+g'(x-4t) \end{align*} and \begin{align*} u_t=4f'(x+4t)-4g'(x-4t) \end{align*} b) We have \begin{align*} u_{xx}=f''(x+4t)+g''(x-4t) \end{align*} and \begin{align*} u_{tt}&=16f''(x+4t)+16g''(x-4t) \\ &=16\left[ f''(x+4t)+g''(x-4t) \right] \\ \end{align*} c) We have \begin{align*} 4\left[f'(x+4t)-g'(x-4t)\right]+4\left[f'(x+4t)+g'(x-4t)\right] = 8f'(x+4t) \end{align*} d) We have \begin{align*} 16\left[ f''(x+4t)+g''(x-4t) \right]-16\left[ f''(x+4t)+g''(x-4t) \right]=0 \end{align*}