For a vector-valued function
[tex]\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))[/tex]
the matrix of partial derivatives (a.k.a. the Jacobian) is the [tex]m\times n[/tex] matrix in which the [tex](i,j)[/tex]-th entry is the derivative of [tex]f_i[/tex] with respect to [tex]x_j[/tex]:
[tex]D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}[/tex]
So we have
(a)
[tex]D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}[/tex]
(b)
[tex]D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}[/tex]
(c)
[tex]Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}[/tex]
(d)
[tex]Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}[/tex]
