I came across an identity for differentiating an unit vector, however I can't prove it so I would appreciate if someone would explain how to derive the identity. Let $f(t)$ be a vector valued function, then its magnitude is given by $||f(t)||$, and $f(t)$ is a differentiable curve such that $f(t) ≠ 0$ for all $t$. Then the derivative of the unit vector is given by $\frac{d}{dt}\frac{f(t)}{||f(t)||}=\frac{f(t)✕f'(t)✕f(t)}{||f(t)||^3}$
Also the unit tangent vector $T(t)$ is defined as: $T(t)=\frac{f'(t)}{||f'(t)||}$ and in the same way $T'(t)=\frac{f'(t)✕f''(t)✕f'(t)}{||f'(t)||}$. I appreciate any help you can provide.
Let $f(t) = (f_1(t), \ldots, f_n(t))$. Since $f(t) / \|f(t)\|$ is a vector, it suffices to take the derivative of $f_i(t) / \|f(t)\|$ for each $i$, and combine all these derivatives into a vector.
It will be useful to note that the chain rule implies $$\frac{d}{dt} \|f(t)\| = \frac{d}{dt} \sqrt{\sum_{j=1}^n f_j(t)^2} = \frac{\sum_{j=1}^n f_j(t) f'_j(t)}{\|f(t)\|}.$$ Then the product rule (you could equivalently use the quotient rule) implies $$\frac{d}{dt} \frac{f_i(t)}{\|f(t)\|} = \frac{f_i'(t)}{\|f(t)\|} - \frac{f_i(t)}{\|f(t)\|^2} \cdot \frac{\sum_{j=1}^n f_j(t) f'_j(t)}{\|f(t)\|} = \frac{f'_i(t) \|f(t)\|^2 - f_i(t) \sum_{j=1}^n f_j(t) f'_j(t)}{\|f(t)\|^3}.$$
This works for any $i$, so the final answer is $$\frac{f'(t) \|f(t)\|^2 - f(t) \sum_{j=1}^n f_j(t) f'_j(t)}{\|f(t)\|^3}$$ where $f(t)$ and $f'(t) := (f'_1(t), \ldots, f'_n(t))$ are vectors. I suppose the cross product in your post should equal the numerator, but I don't feel like going through that right now.