Question

Compute the matrix of partial derivatives of the following functions.

(a) f(x, y) = (ex, sin(xy)) Drx, y) =
(b) f(x, y, z) = (x-y, y + z) Df(x, y, z) =
(c) f(x, y)-(xy, x - y, xy) Df(x, y) =
(d) rx, y, z) = (x + z, y-42, x-y) Df(x, y, z) =

Answer 1

For a vector-valued function

`\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))`

the matrix of partial derivatives (a.k.a. the Jacobian) is the
`m* n` matrix in which the
`(i,j)`-th entry is the derivative of
`f_i` with respect to
`x_j`:

`D\mathbf f(\mathbf x)=\begin{bmatrix}(\partial f_1)/(\partial x_1)&(\partial f_1)/(\partial x_2)&\cdots&(\partial f_1)/(\partial x_n)\\(\partial f_2)/(\partial x_1)&(\partial f_2)/(\partial x_2)&\cdots&(\partial f_2)/(\partial x_n)\\\vdots&\vdots&\ddots&\vdots\\(\partial f_m)/(\partial x_1)&(\partial f_m)/(\partial x_2)&\cdots&(\partial f_n)/(\partial x_n)\end{bmatrix}`

So we have

(a)

`D f(x,y)=\begin{bmatrix}(\partial(e^x))/(\partial x)&(\partial(e^x))/(\partial y)\\(\partial(\sin(xy)))/(\partial x)&(\partial(\sin(xy)))/(\partial y)\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}`

(b)

`D f(x,y,z)=\begin{bmatrix}(\partial(x-y))/(\partial x)&(\partial(x-y))/(\partial y)&(\partial(x-y))/(\partial z)\\(\partial(y+z))/(\partial x)&(\partial(y+z))/(\partial y)&(\partial(y+z))/(\partial z)\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}`

(c)

`Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}`

(d)

`Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}`