数学分析: Partial Derivatives

Partial Derivatives

For function $f(x,y)$ and some point in the domain $(x_0,y_0)\in D$ , we define derivatives of $f(x,y)$ at $(x_0,y_0)$ with respect to $x$ as

f_x(x_0,y_0)=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=\lim_{\Delta x\to 0}\frac{\Delta w}{\Delta x}

which as also be denoted as $\Big( \frac{\partial w}{\partial x} \Big)_{(x_0,y_0)}$

When we extend this to even more variables, the full version looks like this:

f_{x_j}(\mathbf{a})=\lim_{h\to 0}\frac{f(\mathbf{a}+h\mathbf{e}_j)-f(\mathbf{a})}{h}

If $f(\cdot)$ is partially differentiable at every point in $D$ , then we have the partial derivative function $f_x(\cdot)$ , also denoted as $D_{x_j} f(\mathbf{a})$

Corollary: Tangent Plane

我们可以用偏导数计算切于某点的平面，考虑 $f:X\mapsto \mathbb{R}, X\subset \mathbb{R}^2$ ，切于点 $\mathbf{x}=(a,b,f(a,b))$ 的平面是

z=f(a,b)+f_x(a,b)\big(x-a\big)+f_y(a,b)\big(y-b\big)

Example 1. Suppose $z=x^y$ , prove that $\frac{x}{y}\frac{\partial z}{\partial x}+\frac{1}{\ln x}\frac{\partial z}{\partial y}=2z$

Since $z=x^y$ , so

z_x=yx^{y-1}\\ z_y=x^y \ln x

Substituting and we can prove the statement.

Second Order Partial Derivatives

C^k

function

If $f$ ’s $k$ -th derivative exists and is continuous on $D$ , then we denote $f$ as a $C^k$ function.

Proposition: $\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x} \iff$ $f$ is a $C^2$ function

Chain Rule for Partial Derivatives

For vector-valued functions, suppose $\boldsymbol{h}=\boldsymbol{f}\circ\boldsymbol{g}, y_0=g(t_0)$ , then

D\boldsymbol{h}(\boldsymbol{t}_0)=D\boldsymbol{f}(\boldsymbol{y}_0)D\boldsymbol{g}(\boldsymbol{t}_0)

Total Differential

\Delta w\approx dw=f_x(x_0,y_0)dx+f_y(x_0,y_0)dy