## 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.