show that in the one dimensional case, we have show that in the one dimensional case, we have
$\dfrac{\partial \phi}{\partial x}(t,x) = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s,x))ds\right)}$.
I tried to solve using the IVP $x'=f(t,x).x \,\,\,\, , x(t_0)=x_0$ and 
I used $\phi$ is a solution. then replace the solution in the IVP and obtain
$\dfrac{\phi '}{\phi}= f(t,\phi)$. From here, I did the integration of $t_0$ to $t$, then partially derived with respect to $x$. In the end I got this relationship
$\dfrac{\partial \phi}{\partial x} = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s))ds\right)}$.
I'm finding this question very strange because $\phi$ should not be of the form $\phi (t)$? here we have $\phi(t,x)$ in $\dfrac{\partial \phi}{\partial x}(t,x) = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s,x))ds\right)}$. Can someone help?
 A: If you are content with purely formal calculations, you can proceed as follows.  We have
$$
\frac{\partial \phi}{\partial t}(t, x) = f(t, \phi(t, x)),
$$
which, after differentiating in $x$ and changing order of partials, gives
$$
\frac{\partial}{\partial t} \Bigl( \frac{\partial \phi}{\partial x} \Bigr)  (t,x) = \frac{\partial f}{\partial x}(t, \phi(t,x))\, \frac{\partial \phi}{\partial x}(t, x).
$$
So, 
$$
\frac{\partial \phi}{\partial x}  (t,x)
$$
(with $x$ for the moment fixed) satisfies a linear ordinary differential equation
$$
y'(t) = \frac{\partial f}{\partial x}(t, \phi(t,x))\, y(t).
$$
What about the initial value?  One has $\phi(0, x) = x$, so differentiating it in $x$ we get 
$$
y(0) = 1.
$$
Therefore
$$
\dfrac{\partial \phi}{\partial x}(t,x) = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s,x))\,ds\right)}.
$$
Indeed, the above calculations are legitimate, since (assuming that $\partial f/\partial x$ is continuous) $\phi(\cdot, \cdot)$ is so regular that one can change the order of partial differentiations, etc. But that's a different story.
