Regarding the derivative of an implicit line

Question

Let $\sigma(x) = f(x,x^2+1)$, where $f: R^2 \rightarrow R^3$ is of class $C^1$ and

$$Df(0,1) = \begin{bmatrix}0 & 1\\2 & 3\\4 & 5\end{bmatrix}$$

Find $\sigma ' (0)$.

I know that the definition of derivative of a line is

$$\sigma ' (0)=\lim_{h\rightarrow 0} \frac{\sigma(h)-\sigma(0)}{h}$$

Then I can substitute $\sigma$

$$\lim_{h\rightarrow 0} \frac{f(h,h^2+1)-f(0,1)}{h}$$

but this is as far as I get. What should be the next step?

Answer 1

$\sigma$ is a mapping from $R^1$ to $R^3$ so its derivative can be represented by a 3 by 1 matrix (column vector). By the chain rule that B. Pasternak mentioned, $\sigma'= Df(x, y)\begin{bmatrix}1 \\ 2x\end{bmatrix}$. In particular, at "0", x= 0, y= 1, so that is $\sigma'= \begin{bmatrix} 0 & 1 \\ 2 & 3 \\ 4 & 5\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix} 0 \\ 2 \\ 4\end{bmatrix}$

Answer 2

I think you should use the chain rule, as the function $\sigma$ is just a composition $f\circ h$, where $h$ is the function from the real line to $\mathbb{R}^2$ sending $x$ to $(x,x^2+1)$.