Find time derivative of $y(x) = \dfrac{1}{1+\exp(x_2 - x_1) + \exp(x_3-x_1)}$

Question

Let $x: \mathbb{R} \to \mathbb{R}^3$, $x(t) = [x_1(t), x_2(t), x_3(t)]^T$

Let $y:\mathbb{R}^3 \to \mathbb{R}$, $y(x) = \dfrac{1}{1+\exp(x_2 - x_1) + \exp(x_3-x_1)}$

What should be the correct expression for $\dfrac{dy}{dt}$?

Answer 1

The answer of @Shadock is exact, but one can detail a little more.

One must say that derivatives are matrices that are being composed.

If $x: \mathbb{R} \to \mathbb{R}^3$, its derivative matrix will the $1 \times 3$ (line) matrix $[x'_1(t),x'_2(t),x'_3(t)]$.

For $y:\mathbb{R}^3 \to \mathbb{R}$, the derivative will be the column matrix:

$$\begin{bmatrix} \partial y/\partial x_1\\ \partial y/\partial x_2\\ \partial y/\partial x_3 \end{bmatrix} $$

The derivative of the composite function $z=y \circ x:\mathbb{R} \to \mathbb{R}$ is thus the product of these two matrices and the result is naturally a $1 \times 1$ matrix, i.e., a real number.

Thus the result is

$$ z'(t)=(\partial y/\partial x_1)x'_1(t)+(\partial y/\partial x_2)x'_2(t)+(\partial y/\partial x_3)x'_3(t)$$

After calculation, using notation $D=1+e^{x_2-x_1}+e^{x_3-x_1}$,

$$z'(t)=\dfrac{1}{D^2}\left((e^{x_2-x_1}+e^{x_2-x_1})x'_1(t)-e^{x_2-x_1}x'_2(t)-e^{x_3-x_1}x'_3(t)\right)$$