I am learning differential manifold.
Let $f:R\rightarrow R$ by $f(x)=e^{x}$. Find the vector field image of $f\cdot \frac{\partial}{\partial x}$
Here is what I have done so far:
$(f\cdot \frac{\partial}{\partial x})_{x_{0}}$ $=$ $f_{*}(\frac{\partial}{\partial x}$$\mid_{f^{-1}(x_{0})})$ $=$ $f_{*}(\frac{\partial}{\partial x}$$\mid_{logx_{0}})$
Since the Jacobian of $f(x)$ is $e^{x}$, we have the above equation equals $e^{(\frac{\partial}{\partial x}\mid_{logx_{0}})}$.
I don't know how to proceed further. I know the answer is $(x\frac{\partial}{\partial x})_{x_{0}}$ and hence $(f\cdot \frac{\partial}{\partial x})=(x\frac{\partial}{\partial x})$, but I don't know how to get this answer.
Thank you so much for any hints and help!!!!
Okay. I think I was just going nuts. Here is the solution:
Actually everything is right untrue we get the Jacobian of $f(x)$ is $e^{x}$, and thus we have it is $e^{x}\frac{\partial}{\partial x}$$\mid_{log(x_{0})}$, and it is equivalent to the answer.
Moreover, the vector field image is actually just the product of the push-forward to the basis of the vector.