Finding tangent vector for a composite function at a given point I would want to insert the whole question but it's kinda complex/difficult to type it out :( sorry guysss
Basically the problem looks similar to
$T(x,y,z)=(u,v)$
$u=x+y+z^2$
$v=\cos(x)+\tan(y)-z+3$
vector $r(t)=(t\cdot e,(t^2+2)^t, e^{-s^2})$
At point $t=0$ what's the tangent vector of $T(r(t))$?
Anybody who can help me out with this problem?
Would appreciate sooooo much $x$.
 A: We have the functions $T:\mathbb R^3\to \mathbb R^2$ and $r:\mathbb R\to\mathbb R^3$, so the composition is such that $$T\circ r:\mathbb R\overset{r}{\longrightarrow}\mathbb R^3\overset{T}{\longrightarrow} \mathbb R^2$$
$r(0)=(0,1,1)$ and the jacobian matrix of $T$ in $(0,1,1)$ is
$$J_T(0,1,1)=\begin{pmatrix}\nabla u(0,1,1)\\\nabla v(0,0,1) \end{pmatrix}=\begin{pmatrix} 1&&1&&2z\\-\sin(x)&&\dfrac{1}{\cos^2(x)}&&-1 \end{pmatrix}\Bigg|_{(0,1,1)}=\begin{pmatrix}1&&1&&2\\0&&\cos^{-2}(1)&&-1 \end{pmatrix},$$
while the jacobian of $r$ in $0$ is the vector $\begin{pmatrix}e\\\log(2)\\0 \end{pmatrix}$, then for the rule of composition we get $$J_{T\circ r}(0)=\begin{pmatrix}1&&1&&2\\0&&\cos^{-1}(1)&&-1 \end{pmatrix} \cdot\begin{pmatrix}e\\\log(2)\\0 \end{pmatrix}=\begin{pmatrix}e+\log(2) \\\log(2)\cos^{-2}(1) \end{pmatrix}$$
A: Its an application of the Chain rule. Note that $r(0)=(0,1,1)$ and $r'(0)=(e,\ln 2,0)$. By the chain rule
$$\tag{*}\frac{\partial}{\partial t}\biggl|_{t=0}T(r(t))=J(T)_{r(0)}\cdot r'(0)$$
where $J(T)_{r(0)}$ is the Jacobian of $T$ evaluated at $r(0)$. That is,
\begin{align}
J(T)_{r(0)}&=
\begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}& \frac{\partial u}{\partial z} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z}
\end{pmatrix}
_{\bigl|r(0)}\\
\\
&=\begin{pmatrix}
1 & 1 & 2z\\
-\sin x & \frac{1}{\cos ^2 y} & -1
\end{pmatrix}
_{(0,1,1)}\\
\\
&=\begin{pmatrix}
1 & 1 & 2\\
0& \frac{1}{\cos^2 1} & -1
\end{pmatrix}
\end{align}
Multiplying the above matrix with the vector $r'(0)=(e,\ln 2,0)$ and using $(*)$ we obtain
\begin{align}
\frac{\partial}{\partial t}\biggl|_{t=0}T(r(t))&=\begin{pmatrix}
1 & 1 & 2\\
0& \frac{1}{\cos^2 1} & -1
\end{pmatrix} \cdot 
\begin{pmatrix}
e \\ \ln 2 \\ 0
\end{pmatrix}\\
&\\
&=(e+\ln 2,\  \frac{\ln 2}{\cos^2 1})
\end{align}
the desired tangent vector of $T(r(t))$ at $t=0$.
