4.3 Partial derivatives (Page 2/11)

Calculus volume 3 Page 2 / 11

Use the definition of the partial derivative as a limit to calculate $\partial f / \partial x$ and $\partial f / \partial y$ for the function

f (x, y) = 4 x^{2} + 2 x y - y^{2} + 3 x - 2 y + 5 .

$\frac{\partial f}{\partial x} = 8 x + 2 y + 3, \frac{\partial f}{\partial y} = 2 x - 2 y - 2$

Got questions? Get instant answers now!

The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix $y$ and define $g (x) = f (x, y)$ as a function of $x .$ Then

g^{'} (x) = lim_{h \to 0} \frac{g (x + h) - g (x)}{h} = lim_{h \to 0} \frac{f (x + h, y) - f (x, y)}{h} = \frac{\partial f}{\partial x} .

The same is true for calculating the partial derivative of $f$ with respect to $y .$ This time, fix $x$ and define $h (y) = f (x, y)$ as a function of $y .$ Then

h^{'} (x) = lim_{k \to 0} \frac{h (x + k) - h (x)}{k} = lim_{k \to 0} \frac{f (x, y + k) - f (x, y)}{k} = \frac{\partial f}{\partial y} .

All differentiation rules from Introduction to Derivatives apply.

Calculating partial derivatives

Calculate $\partial f / \partial x$ and $\partial f / \partial y$ for the following functions by holding the opposite variable constant then differentiating:

$f (x, y) = x^{2} - 3 x y + 2 y^{2} - 4 x + 5 y - 12$
$g (x, y) = sin (x^{2} y - 2 x + 4)$

To calculate $\partial f / \partial x,$ treat the variable $y$ as a constant. Then differentiate $f (x, y)$ with respect to $x$ using the sum, difference, and power rules:

$\begin{matrix} \frac{\partial f}{\partial x} & = \frac{\partial}{\partial x} [x^{2} - 3 x y + 2 y^{2} - 4 x + 5 y - 12] \\ = \frac{\partial}{\partial x} [x^{2}] - \frac{\partial}{\partial x} [3 x y] + \frac{\partial}{\partial x} [2 y^{2}] - \frac{\partial}{\partial x} [4 x] + \frac{\partial}{\partial x} [5 y] - \frac{\partial}{\partial x} [12] \\ = 2 x - 3 y + 0 - 4 + 0 - 0 \\ = 2 x - 3 y - 4. \end{matrix}$

The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable $x,$ so they are treated as constant terms. The derivative of the second term is equal to the coefficient of $x,$ which is $−3 y .$ Calculating $\partial f / \partial y :$

$\begin{matrix} \frac{\partial f}{\partial y} & = \frac{\partial}{\partial y} [x^{2} - 3 x y + 2 y^{2} - 4 x + 5 y - 12] \\ = \frac{\partial}{\partial y} [x^{2}] - \frac{\partial}{\partial y} [3 x y] + \frac{\partial}{\partial y} [2 y^{2}] - \frac{\partial}{\partial y} [4 x] + \frac{\partial}{\partial y} [5 y] - \frac{\partial}{\partial y} [12] \\ = −3 x + 4 y - 0 + 5 - 0 \\ = −3 x + 4 y + 5. \end{matrix}$

These are the same answers obtained in [link] .
To calculate $\partial g / \partial x,$ treat the variable y as a constant. Then differentiate $g (x, y)$ with respect to x using the chain rule and power rule:

$\begin{matrix} \frac{\partial g}{\partial x} & = \frac{\partial}{\partial x} [sin (x^{2} y - 2 x + 4)] \\ = cos (x^{2} y - 2 x + 4) \frac{\partial}{\partial x} [x^{2} y - 2 x + 4] \\ = (2 x y - 2) cos (x^{2} y - 2 x + 4) . \end{matrix}$

To calculate $\partial g / \partial y,$ treat the variable $x$ as a constant. Then differentiate $g (x, y)$ with respect to $y$ using the chain rule and power rule:

$\begin{matrix} \frac{\partial g}{\partial y} & = \frac{\partial}{\partial y} [sin (x^{2} y - 2 x + 4)] \\ = cos (x^{2} y - 2 x + 4) \frac{\partial}{\partial y} [x^{2} y - 2 x + 4] \\ = x^{2} cos (x^{2} y - 2 x + 4) . \end{matrix}$

Got questions? Get instant answers now!

Calculate $\partial f / \partial x$ and $\partial f / \partial y$ for the function $f (x, y) = tan (x^{3} - 3 x^{2} y^{2} + 2 y^{4})$ by holding the opposite variable constant, then differentiating.

$\begin{matrix} \frac{\partial f}{\partial x} = (3 x^{2} - 6 x y^{2}) {sec}^{2} (x^{3} - 3 x^{2} y^{2} + 2 y^{4}) \\ \frac{\partial f}{\partial y} = (−6 x^{2} y + 8 y^{3}) {sec}^{2} (x^{3} - 3 x^{2} y^{2} + 2 y^{4}) \end{matrix}$

Got questions? Get instant answers now!

How can we interpret these partial derivatives? Recall that the graph of a function of two variables is a surface in $ℝ^{3} .$ If we remove the limit from the definition of the partial derivative with respect to $x,$ the difference quotient remains:

\frac{f (x + h, y) - f (x, y)}{h} .

This resembles the difference quotient for the derivative of a function of one variable, except for the presence of the $y$ variable. [link] illustrates a surface described by an arbitrary function $z = f (x, y) .$

A complicated curve in xyz space with a secant line through the points (x, y, f(x, y)) and (x + h, y, f(x + h, y)). — Secant line passing through the points $(x, y, f (x, y))$ and $(x + h, y, f (x + h, y)) .$

In [link] , the value of $h$ is positive. If we graph $f (x, y)$ and $f (x + h, y)$ for an arbitrary point $(x, y),$ then the slope of the secant line passing through these two points is given by

\frac{f (x + h, y) - f (x, y)}{h} .

This line is parallel to the $x -axis.$ Therefore, the slope of the secant line represents an average rate of change of the function $f$ as we travel parallel to the $x -axis.$ As $h$ approaches zero, the slope of the secant line approaches the slope of the tangent line.

<< Chapter < Page Page > Chapter >>

Practice Key Terms 4

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask

©flickr: Gareth	Resume Writing MCQ By Abby Sharp Start Quiz
	1 AP 01 Human Body Anatomy Physiology MCQ By OpenStax Start Quiz
	26 AP 26 Fluid, Electrolyte, Acid-Base Balance Essay By OpenStax Start Flashcards
	Chemistry Practice Test By Sandhills MLT Start Test
	1 Neuroscience Exam 2004 1 By David Corey Start Exam
	23 AP Key Terms 23 The Digestive System By OpenStax Start Key Terms
	11 AP 11 Muscular System MCQ By OpenStax Start Quiz
	1 Gastrointestinal Pathophysiology By Laurence Bailen Start Exam
	Advertising Promotion BUS210 By Melinda Salzer Start Quiz
	19 AP 19 Cardiovascular System Heart Essay By OpenStax Start Flashcards