Partial Derivative of $G(t,x(t)) = e^{-pt} \frac{u_x(x(t))}{u_x(x(0))}$

Question

I have given the following function for some constant $p$ and function $u(x(t))$

$$G(t,x(t)) = e^{-pt} \frac{u_x(x(t))}{u_x(x(0))}$$

where $$ u_x(x(t)) = \frac{\partial u(x(t))}{\partial x} $$

Now I want to calculate

$$\frac{\partial G(t,x(t)) }{\partial x} $$

But I just cannot wrap my head around it. Better said I don't know how to start to write it down - any help would be appreciated!

Answer 1

So you are considering a function $G$ of the vector (presumably): $(t, x).$ So wherever $x(t)$ appears, change it to $x$ and remark that $u_x(x(0))$ is simply a constant, say $c,$ so your function $G$ takes the form $G(t, x) = c^{-1} e^{-pt} u'(x),$ where $u$ is a function of $x$ alone and hence, $u'$ makes sense. The partial derivative of $G$ would be simply $G_x(t, x) = c^{-1} e^{-pt} u''(x) = c^{-1} e^{-pt} u_{xx}(x).$ Hence, $G(t, x(t)) = e^{-pt} \dfrac{u_{xx}(x(t))}{u_x(x(0))}.$

Answer 2

Ok my solution would be to treat $u_x(x(0))$ as a constant because $x(0)=1$ and thus

$$\frac{\partial G(t,x(t)) }{\partial x} = e^{-pt} \frac{u_{xx}(x(t))}{u_x(x(0))} $$