Imagine any curve $y=f(x)$ in a cartesian coordinate system. At any point, A vector along the tangent can be given as $$ \vec V = \hat i + \frac{dy} {dx} \hat j $$ I'm trying to find the direction of $d\vec V$ vector, i.e. : $\vec V(x+dx) - \vec V(x)$
$$ d\vec V = \left(\hat i+\left.\frac{dy}{dx}\right|_{x+dx} \hat j\right) - \left(\hat i+\left.\frac{dy}{dx}\right|_{x} \hat j\right)=\frac{d^2y}{dx^2}dx\hat j $$
I doubt this result because if I turn my coordinate axes, keeping the curve as such, the vector, mathematically now orients along my new $\hat j'$ axis. As explained in the diagram:
Please explain the mistakes in my approach and provide a method to find the direction of $d \vec V$.
I don't know the math terms used while dealing with vectors like Fields and manifolds. Please understand the situation physically. The components of the vector are obviously real numbers and the vector is 2-dimensional.
To spell out what is mentioned in the comments:
Consider the unit circle parametrized by $x=t$, $y=(1-t^2)^{1/2}$ for $t\in[-1,1]$, where $(x,y)$ are standard Cartesian coordinates. Note that the acceleration vector is vertical everywhere. ("Vertical" here means having standard $x$ coordinate zero.)
Now rotate your axes 45 degrees counterclockwise, and let the new coordinates be $(x', y')$.
Thus $x'=2^{-1/2}(x+y)$, $y'=2^{-1/2}(y-x)$.
So, the point $x=t$, $y=(1-t^2)^{1/2}$ corresponds to the point
$$x'=2^{-1/2}\big(t+(1-t^2)^{-1/2}\big), y'=2^{-1/2}\big((1-t^2)^{-1/2}-t\big)$$
You can clearly see that the acceleration vector of this parametrized curve will not have $x'$ coordinate identically equal to zero. In other words, the acceleration vector is not "vertical in the new coordinates."