## DESCRIPTION
## This is Problem 2.5.33 from the APEX Calculus text. It covers a basic application of the Quotient Rule.
## ENDDESCRIPTION
## DBsubject(Calculus - single variable)
## DBchapter(Differentiation)
## DBsection(Derivatives of logarithmic functions)
## Level(2)
## Institution('Valdosta State University')
## Author('S. V. Ault')
## RevisedBy('F. J. Francis')
## TitleText1('APEX Calculus')
## AuthorText1('Hartman')
## EditionText1('3.0')
## Section1('2.5')
## Problem1('33')
## MO(1)
## Keywords('derivative', 'logarithms', 'chain rule', 'ULETH-MATH1560', 'ULETH-MATH1565')
###########################
# Initialization
DOCUMENT();
loadMacros(
"PGstandard.pl",
"MathObjects.pl",
# Allows a single problem to contain multiple parts, where later
# parts aren't visible until the earlier ones are completed.
"scaffold.pl",
# Provides greater control over the layout of the problem.
"PGML.pl",
# Used for course-specific initializations.
"PGcourse.pl",
# Used to provide contextual help for how to type solutions.
"AnswerFormatHelp.pl",
# For the popup
"parserPopUp.pl"
);
TEXT(beginproblem());
###################################
# Setup
#-ULETH-#
Context("Numeric")->variables->add(k => 'Real');
Context()->variables->add(u => 'Real');
$ans1 = Formula("ln(u)");
$ans2 = Formula("kx");
$ans3 = Formula("ln(k)");
$ans = Formula("1/x");
$op = PopUp(["?", "+", "-", "*", "/"], "+");
#-ENDULETH-#
###################################
# Main Text
#-ULETH-#
# Initial problem text to explain the scaffolding setup.
BEGIN_PGML
Let [`k`] be a constant.
Compute [`\frac{d}{dx}[ \ln (kx) ]`] in two ways.
This problem has two parts. You may only open the next part after correctly
completing the previous part.
END_PGML
#-ENDULETH-#
Scaffold::Begin();
#-ULETH-#
# For each part the answer or answer checker reference must appear in the curly
# braces immediately after the answer field.
Section::Begin("Chain Rule");
# Part 1 problem text.
BEGIN_PGML
Using the Chain Rule, first decompose [`y = \ln(kx)`]
into an outside and inside function.
Outside function (in terms of [`u`]): [`\quad y`] = [__________]{Compute($ans1)}
Inside function (in terms of [`x`]): [`\quad u`] = [__________]{Compute($ans2)}
Then find the derivative, [``\frac{d}{dx}[ \ln (kx) ] = ``][___________]{Compute($ans)}
END_PGML
Section::End();
Section::Begin("Algebraic Manipulation");
# Part 2 problem text.
BEGIN_PGML
Using a law of logarithms to simplify first:
[` \ln(kx) = `][____]{Compute($ans3)} [____]{$op} [`\, \ln x `].
(Fill in the blanks to make this a true statement.)
Now take the derivative of the simplified function:
[`\frac{dy}{dx} [ \ln(kx) ]`]= [_____________________]{Compute($ans)}
END_PGML
Section::End();
#-ENDULETH-#
Scaffold::End();
###################################
# Answer Evaluation
#evaluation must be completed in the problem text
###################################
# Solution
BEGIN_PGML_SOLUTION
*Chain Rule*
The decomposition is [` y = \ln u`], and [`u = kx`]. By
the chain rule,
[`\displaystyle
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
= \frac{1}{u} \cdot k
= \frac{1}{kx} \cdot k
= \frac{1}{x}.
`]
*Algebraic Manipulation*
After simplifying [` \ln(kx) = \ln k + \ln x`], take the derivative
(note, [`\ln k`] is simply a constant).
[`\displaystyle
\frac{dy}{dx} = 0 + \frac{1}{x} = \frac{1}{x}.
`]
The answers in both parts should match.
END_PGML_SOLUTION
COMMENT("This is a multi-part problem
in which the next part is revealed only after the previous
part is correct.
There is no randomization in this question.
Includes a solution set.
Made from a ULETH template.
");
#-ENDULETH-#
ENDDOCUMENT();