Let $\gamma:[0,2\pi] \to \Bbb R^3$ be $$\gamma(t)=(4t,\cos(3t),\sin(3t)).$$
Justify that $\gamma$ is differentiable.
This is a question of a past Analysis exam and I do not know how to go about it. All the information I have found in the internet is about finding a derivative of a parametric function whose codomain is $\Bbb R^2$.
My question is: How do I show that this function is differentiable?
Thanks in advance.
A vector valued function $f:\mathbb{R}^n\to\mathbb{R}^m$ is differentiable if its component functions $f_i:\mathbb{R}^n\to \mathbb{R}$ are differentiable. Here, $n=1$ and $m=3$. The component functions are,
$$f_1(t) = 4t\\ f_2(t)=\cos(3t)\\ f_3(t)=\sin(3t) $$
Hopefully it is clear that each of these component functions are differentiable. Then, we have that $f$ is differentiable and its derivative is the Jacobian (matrix of partial derivatives),
$$ \nabla f = \begin{pmatrix} \partial_t f_1 \\ \partial_t f_2 \\ \partial_t f_3 \end{pmatrix} = \begin{pmatrix} 4 \\ -3\sin(3t) \\ 3\cos(3t) \end{pmatrix} $$