Partial Derivatives

Figure 4.13: $z = f(x,y)$

Understanding

A partial derivative of $f(x,y)$ at $(x_0, y_0)$ with respect to $x$ is a curve shows up by slicing the surface given by $z = f(x,y)$ by the plane $y = y_0$. Thus differentiate $z = f(x,y_0)$ by $x$ gives the answer.

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

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

is called coefficient of partial derivative with respect to $x$. Similarly, Keep $x = x_{0}$ a constant and a function $f(x_0,y)$ of $y$ is differentiable at $y = y_{0}$. Then

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

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 $f(x,y)$ is differentiable at esch $y$ in $D$, then $f(x,y)$ is called partially differentiable on $D$ with respect $y$.

Partially Differentiable

$f(x,y)$ is differentiable with respect to $x$ at $(x_0, y_0)$ if and only if $f(x,y_0)$ is differentiable at $x=x_0$. Similarly, $f(x,y)$ is differentiable with respect to $y$ at $(x_0, y_0)$ if and only if $f(x_0,y)$ is differentiable at $y = y_0$.

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

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

$f(x,1)$ is a curve given by slicing the surface $z = f(x,y)$ with the plane $y = 1$. Then $f_{x}(1,1)$ is a derivative of the curve $z = f(x,1)$ with respect to $x$ at $x = 1$.

Figure 4.14: Example4-6

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 $f(x,1)$ by $x$. Then we have $f_x(x,1) = 2x$. Thus, $f_x(1,1) = 2$ $\blacksquare$

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

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

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, $f_y(1,1) = \frac{1}{2}$ $\blacksquare$

Figure 4.15: coefficients

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

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

Similarly for a partial derivative of $f(x,y)$ with respect to $y$ is

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

Finding a partial derivative, we say partial differentiation

$f_{y}(1,1)$ is the derivative $f(y,1)$ with respect to $y$ at $y = 1$.

Figure 4.16: Exercise4-6

Understanding

To find the partial derivative of $f(x,y)$ with respect to $x$, fix $y$ and differentiate $f(x,y)$ by $x$. $\partial$ is read del, dee, partial.

Example 4..7   Find the partial derivative of the followings.
1. $f(x,y) = x^2 + 2xy + 3y^3$2. $g(x,y) = \tan^{-1}(\frac{y}{x})$

Note that to find the partial derivative of $f(x,y)$ with respect to $y$, we treat $x$ as constant. Thus $\partial{(x^2)}/{\partial y} = 0$.

SOLUTION 1. To find $f_{x}(x,y)$, treat $y$ as constant and differentiate $f(x,y)$ with respect ot $x$.

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

$$= 2x + 2y$$

Also, treat $x$ as constant and differentiate with respect to $y$

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

$$= 2x + 9y^2 \ensuremath{ \blacksquare}$$

Review

1. $(t^{\alpha})' = \alpha t^{\alpha - 1}$
2. $(e^{t})' = e^t$
3. $(\log\vert t\vert)' = \frac{1}{t}$
4. $(\sin{t})' = \cos{t}$
5. $(\cos{t})' = -\sin{t}$
6. $(\tan{t})' = \frac{1}{\cos^{2}{t}}$
7. $(\sin^{-1}{t})'=\frac{1}{\sqrt{1-t^2}}$
8. $(\tan^{-1}{t})' = \frac{1}{1+t^2}$

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

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

$$= \frac{1}{1+t^2}\cdot (-\frac{y}{x^2}) = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{-y}{x^2} = \frac{-y}{x^2 + y^2}$$

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

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

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

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

$$\frac{\partial z}{\partial x} = \frac{\partial (\sqrt{t})}{\partial t}\cdot \frac{\partial t}{\partial x} = \frac{1}{2\sqrt{t}}\cdot 2x = \frac{x}{\sqrt{x^2 + y^2}}$$

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

$(\sqrt{t})' = (t^{\frac{1}{2}})' = \frac{1}{2}t^{-\frac{1}{2}} = \frac{1}{2\sqrt{t}}$
$(\frac{1}{x})' = (x^{-1})' = -x^{-2} = -\frac{1}{x^2}$

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

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

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

$(\log{t})' = \frac{1}{t}$

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.

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

$f_{x}(x,y) = \frac{y(x^2 + y^2) - xy(2x)}{(x^2 + y^2)^2} = \frac{y^3 - x^2 y}{(x^2 + y^2)^2}$. Now substitute $(0,0)$ to get $f_{x}(0,0) = \frac{0}{0}$. Then conclude that $f(x,y)$ is not partially differentiable is not right. The reason is that $f_{x}(x,y)$ is not continuous at $(0,0)$. Thus substituting $(0,0)$ is not allowed.

SOLUTION To find $f_{x}(0,0)$, substitute $y = 0$ and find $f(x,0)$. Then

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

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.

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

To find the function $f(x,y)$ is differentiable with respect to $x$ at $(0,0)$, it is enough to check $f{(x,0)}$ is differentiable with respect to $x$.

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

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

which implies that $f_{y}(0,0) = \frac{0}{1} = 0$. Thus $f(x,y)$ is partially differentiable with respect to $x$ and $y$ at $(0,0)$ $\blacksquare$

Exercise

1.

Find the partial derivative of.the following functions.

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

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

2.

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

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

(c) $${f(x,y) = \sin(3x - 2y)}$$ (d) $${f(x,y) = xe^{2y}}$$

Exercise B

1.

Find the partial derivative of the following functions.

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

2.

Find all 2nd partial derivatives of the following functions.

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

(c) $${z = \tan^{-1}{(x^2 + y^2)}}$$

3.

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

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