In the following definition, recall the standard basis vectors of \(\mathbb{R}^n\): \[ \mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \quad \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \quad \ldots, \quad \mathbf{e}_n = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix}. \]
Suppose we write the variables in \(\mathbb{R}^n\) as \(x_1,x_2,\ldots,x_n\), so a function \(f:\mathbb{R}^n \to \mathbb{R}\) takes the form \(z = f(x_1,x_2,\ldots,x_n)\).
Definition.
Let \(f:\mathbb{R}^n \to \mathbb{R}\) be a differentiable function and \((a_1,a_2,\ldots,a_n)\) a fixed point in \(\mathbb{R}^n\).
How do you compute partial derivatives?
To compute the partial derivative of a function \(f:\mathbb{R}^n \to \mathbb{R}\) with respect to \(x_j\), we treat all other variables as constants and differentiate with respect to \(x_j\) using the usual rules of differentiation.
Compute all partial derivatives of the following functions:
Theorem.
Suppose that \(f: \mathbb{R}^n\to \mathbb{R}\) is a differentiable function. Then, as a \(1\times n\) matrix, we have \[ f'(x_1,x_2,\ldots,x_n) = \begin{bmatrix} \displaystyle\frac{\partial f}{\partial x_1}(x_1,x_2,\ldots,x_n) & \displaystyle\frac{\partial f}{\partial x_2}(x_1,x_2,\ldots,x_n) & \cdots & \displaystyle\frac{\partial f}{\partial x_n}(x_1,x_2,\ldots,x_n) \end{bmatrix}. \]
Theorem.
Suppose that \(f: \mathbb{R}^n\to \mathbb{R}\) is a differentiable function and \((a_1,a_2,\ldots,a_n)\) is a fixed point in \(\mathbb{R}^n\). Then, the tangent approximation to the function at this point is given by
\[ L(x_1,x_2,\ldots,x_n) = \sum_{j=1}^n \frac{\partial f}{\partial x_j}(a_1,a_2,\ldots,a_n)(x_j-a_j) + f(a_1,a_2,\ldots,a_n). \]
(a) Computing derivatives
Compute the (total) derivatives of the following functions:
(b) Computing tangent approximations
Comput the tangent approximation of the function \(f\) in part (a) at the point \((1,1,1)\).
If \(f: \mathbb{R}^n \to \mathbb{R}\) is a differentiable function, then we have the partial derivative functions \[ \frac{\partial f}{\partial x_j}: \mathbb{R}^n \to \mathbb{R}, \quad j=1,2,\ldots,n. \]
These functions may themselves be differentiable, in which case we can take their partial derivatives.
Definition.
Let \(f:\mathbb{R}^n \to \mathbb{R}\) be a differentiable function. The second-order partial derivatives of \(f\) are the partial derivatives of the first-order partial derivatives, i.e. \[ \frac{\partial^2 f}{\partial x_i \partial x_j} = \frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right). \]
These may or may not exist.
A mixed second-order partial derivative is a second-order partial derivative of the form \(\displaystyle\frac{\partial^2 f}{\partial x_i \partial x_j}\) with \(i \neq j\).
A pure second-order partial derivative is a second-order partial derivative of the form \(\displaystyle\frac{\partial^2 f}{\partial x_j^2}\).
Theorem (Equality of mixed second-order partial derivatives).
Suppose that \(f:\mathbb{R}^n \to \mathbb{R}\) is a function whose second-order partial derivatives are continuous. Then, for all \(i,j = 1,2,\ldots,n\), we have
\[ \frac{\partial^2 f}{\partial x_i \partial x_j} = \frac{\partial^2 f}{\partial x_j \partial x_i}. \]
Compute all third-order partial derivatives of the function
\[ f: \mathbb{R}^2 \to \mathbb{R}, \quad f(x,y) = 3xy^2-2y+5x^2y^2. \]