What's the tangent space of $M=\{(x,x^3,e^{x-1}): x \in \Bbb{R}\}$ at the point $(1,1,1)$, where $M$ is a manifold of smoothness $C^\infty$.
I know how to find the tangent space of a manifold in the form that gives an implicit function such as $M=\{(x,y,z) \in \Bbb{R}^3: x^2+y^2-z^2=1\}$. The tangent space of $M$ in this case $= \ker(\mbox{dg}(x))$ at the given point which as $2x+zy-2z=0$.
Can anyone help with the question that only the coordinate was given? Any hint would be helpful. :)
From the comment above by @mastrok.
Your manifold is just a one-dimensional curve, so the tangent space should be the tangent line at that point. So, it is the set of points of the form $$\{ (p,v) : p = (1,1,1) \text{ and } v = (\lambda, 3\lambda, \lambda), \lambda \in \Bbb{R} \}.$$