Partial Derivatives

**Figure 4.1:**
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Consider the limit of a function $z = f(\boldsymbol{x}) = f(x,y)$ at point $\boldsymbol{x}_{0} = (x_{0},y_{0})$ . Let $y = y_{0}$ be a constant and a function $f(x,y_{0})$ of $x$ is differentiable at $x = x_{0}$ . Then

$\displaystyle f_{x}(x_0,y_0) = \lim_{h \rightarrow 0}\frac{f(x_{0}+h,y_{0}) - f(x_{0},y_{0})}{h}$

is called coefficient of partial derivative with respect to $x$

. Similarly, Keep $x = x_{0}$ a constant and a function $f(x_0,y)$

is differentiable at $y = y_{0}$ . Then

$\displaystyle f_{y}(x_0,y_0) = \lim_{k \rightarrow 0}\frac{f(x_{0},y_{0}+k) - f(x_{0},y_{0})}{k}$

is called partial differential coefficient with respect to $y$

NOTE If $f_{x}(x_0,y_0)$ exists, then $f(x,y)$ is called partially differentiable with respect $x$ at $(x_0, y_0)$ . If is differentiable at esch $y$ in $D$ , then is called partially differentiable on $D$ with respect $y$ .

Example 4..6 Find the partial differential coefficient of the following function at $(1,1)$

$\displaystyle f(x,y) = x^{2} + y^{2}$

SOLUTION To slice the surface $f(x,y) = x^2 + y^2$

by the plane $y = 1$

, it is enough to consider $f(x,1) = x^2 + 1$

. Now differentiate $f(x,1)$

. Then we have $f_x(x,1) = 2x$

. Thus,

$\blacksquare$

Exercise 4..6 Find the partial differential coefficient of the following function at $(1,1)$

$\displaystyle f(x,y) = \sqrt{xy}$

SOLUTION Let $x = 1$

. Then slice the surface $f(x,y) = \sqrt{xy}$ by the plane $x = 1$

to get the curve $f(1,y) = \sqrt{y}$ . Now differentiate $f(1,y)$

with respect to $y$

to obtain $f_y(1,y) = \frac{1}{2\sqrt{y}}$ . Thus, $f_y(1,1) = \frac{1}{2}$ $\blacksquare$

A partial derivative of $f(x,y)$ with respect to $x$ is the derivative of by keeping $y$ as constant. More precisely,

$\displaystyle \frac{\partial f(x,y)}{\partial x} = f_{x}(x,y) = \lim_{h \rightarrow 0}\frac{f(x+h,y) - f(x,y)}{h}$

Similarly for a partial derivative of $f(x,y)$

with respect to $y$

$\displaystyle \frac{\partial f(x,y)}{\partial y} = f_{y}(x,y) = \lim_{k \rightarrow 0}\frac{f(x,y+k) - f(x,y)}{k}$

Finding a partial derivative, we say partial differentiation

Example 4..7 Find the partial derivative of the followings.
1. $f(x,y) = x^2 + 2xy + 3y^3$

2. $g(x,y) = \tan^{-1}(\frac{y}{x})$

SOLUTION 1. To find $f_{x}(x,y)$ , treat $y$

as constant and differentiate $f(x,y)$

with respect ot $x$

$\displaystyle f_{x}(x,y)$	$\displaystyle =$	$\displaystyle \frac{\partial (x^2)}{\partial x} + \frac{\partial(2xy)}{\partial x} + \frac{\partial (3y^3)}{\partial x}$
	$\displaystyle =$	$\displaystyle 2x + 2y$

Also, treat $x$

as constant and differentiate with respect to $y$

$\displaystyle f_{y}(x,y)$	$\displaystyle =$	$\displaystyle \frac{\partial (x^2)}{\partial y} + \frac{\partial(2xy)}{\partial y} + \frac{\partial (3y^3)}{\partial y}$
	$\displaystyle =$	$\displaystyle 2x + 9y^2 \ensuremath{ \blacksquare}$

2. Let $t = \frac{y}{x}$ . Then

$\displaystyle g_{x}(x,y)$	$\displaystyle =$	$\displaystyle \frac{\partial (\tan^{-1}{t})}{\partial x} = \frac{\partial(\tan^{-1}{t})}{\partial t}\cdot \frac{\partial t}{\partial x}$
	$\displaystyle =$	$\displaystyle \frac{1}{1+t^2}\cdot (-\frac{y}{x^2}) = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{-y}{x^2} = \frac{-y}{x^2 + y^2}$

$\displaystyle g_{y}(x,y)$	$\displaystyle =$	$\displaystyle \frac{\partial(\tan^{-1}{t})}{\partial y} = \frac{\partial(\tan^{-1}{t}))}{\partial t}\cdot \frac{\partial t}{\partial y}$
	$\displaystyle =$	$\displaystyle \frac{1}{1+t^2}\cdot (\frac{1}{x}) = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{x}{x^2} = \frac{x}{x^2 + y^2} \ensuremath{ \blacksquare}$

Exercise 4..7 Find the partial derivatives of the followings.
1. $\displaystyle{z = \sqrt{(x^2 + y^2)}}$ 2. $\displaystyle{z = \log{(x^2 + y^2)}}$

SOLUTION 1. Let $t = x^2 + y^2$ . Then

$\displaystyle \frac{\partial z}{\partial x}$	$\displaystyle =$	$\displaystyle \frac{\partial (\sqrt{t})}{\partial t}\cdot \frac{\partial t}{\partial x} = \frac{1}{2\sqrt{t}}\cdot 2x = \frac{x}{\sqrt{x^2 + y^2}}$
$\displaystyle \frac{\partial z}{\partial y}$	$\displaystyle =$	$\displaystyle \frac{\partial (\sqrt{t})}{\partial t}\cdot \frac{\partial t}{\pa... ...c{1}{2\sqrt{t}}\cdot 2y = \frac{y}{\sqrt{x^2 + y^2}}\ensuremath{ \blacksquare}$

2. Let

. Then

$\displaystyle \frac{\partial z}{\partial x}$	$\displaystyle =$	$\displaystyle \frac{\partial (\log{t})}{\partial t}\cdot \frac{\partial t}{\partial x} = \frac{1}{t}\cdot 2x = \frac{2x}{x^2 + y^2}$
$\displaystyle \frac{\partial z}{\partial y}$	$\displaystyle =$	$\displaystyle \frac{\partial (\log{t})}{\partial t}\cdot \frac{\partial t}{\partial y} = \frac{1}{t}\cdot 2y = \frac{2y}{x^2 + y^2}\ensuremath{ \blacksquare}$

Example 4..8 Find the $f_{x}(0,0), f_{y}(0,0)$ . Then determine the following function is partially differentiable with respect to $x,y$

or not.

$\displaystyle f(x,y) = \left\{\begin{array}{cl} \frac{xy}{x^2 + y^2} & (x,y) \neq (0,0)\\ 0 & (x,y) = (0,0) \end{array} \right.$

SOLUTION To find $f_{x}(0,0)$ , substitute $y = 0$

and find

. Then

$\displaystyle f(x,0) = \frac{x\cdot 0}{x^2 + 0^2} = \frac{0}{x^2} = 0$

Then $f_{x}(x,0) = 0$ . Thus, $f_{x}(0,0) = 0.$

Since $f(0,y) = \frac{0 \cdot y}{0^2 + y^2} = \frac{0}{y^2} = 0$ , we have $f_{y}(0,y) = 0.$ Thus $f_{y}(0,0) = 0$ .

Therefore, $f(x,y)$ is partially differentiable with respect to $x,y$ at $(0,0)$ $\blacksquare$

Exercise 4..8 Find the $f_{x}(0,0), f_{y}(0,0)$ . Then determine the following function is partially differentiable with respect to $x,y$

or not.

$\displaystyle f(x,y) = \log{(1 + xy + y^2)}$

SOLUTION Since $f(x,0) = \log(1 + 0 + 0) = \log{1} = 0$ , $f_{x}(x,0) = 0$ and $f_{x}(0,0) = 0$ .

Note that $f(0,y) = \log(1 + 0 + y^2) = \log(1+y^2)$ . Thus

$\displaystyle f_{y}(0,y) = \frac{2y}{1+y^2}.$

which implies that $f_{y}(0,0) = \frac{0}{1} = 0$ . Thus $f(x,y)$

is partially differentiable with respect to $x$

and

$\blacksquare$

Exercise

1.

Find the partial derivative of.the following functions.

(a) $\displaystyle{f(x,y) = 3x^{2} - xy + y}$ (b) $\displaystyle{f(x,y) = x^{2}e^{-y}}$ (c) $\displaystyle{z = \sqrt{(x^2 + y^2)}}$

(d) $\displaystyle{z = x\sin{y}}$ (e) $\displaystyle{z = \frac{x-y}{x+y}}$

2.

Find all 2nd partial derivatives of the following functions.．

(a) $\displaystyle{f(x,y) = ax^{2} + 2bxy + cy^{2}}$ (b) $\displaystyle{f(x,y,z) = (x+y^{2}+z^{3})^{2}}$

Exercise B

1.

Find the partial derivative of the following functions.

(a) $\displaystyle{z = x^3 + xy^2 + y^3}$ (b) $\displaystyle{z = e^{x} \sin{y}}$ (c) $\displaystyle{z = \log{(x^2 + y^2)}}$

2.

Find all 2nd partial derivatives of the following functions.

(a) $\displaystyle{z = x^3 y + x y^2}$ (b) $\displaystyle{z = x y^2 e^{\frac{x}{y}}}$

3.

Determine the following functions are partially differentiable at the origin.．

(a) $\displaystyle{f(x,y) = \left\{\begin{array}{cl} \frac{y^3 - x^2 y}{x^2 + y^2}, & (x,y) \neq (0,0)\\ 0, & (x,y) = (0,0) \end{array}\right.}$ (b) $f(x,y) = \log{(1 + xy + y^2)}$