** Using WeBWorK **

## DESCRIPTION

##   The Chain Rule

## ENDDESCRIPTION

## Tagged by nhamblet

## DBsubject(Calculus - single variable)

## DBchapter(Differentiation)

## DBsection(Chain rule (without trigonometric functions))

## Institution(Rochester)

## Level(2)

## MO(1)

## KEYWORDS('Derivative', 'Polynomial', 'Chain')

## Library/Rochester/setDerivatives2Formulas/c2s5p3

DOCUMENT();        # This should be the first executable line in the problem.

loadMacros("PG.pl",

           "PGbasicmacros.pl",

           "PGchoicemacros.pl",

           "PGanswermacros.pl",

           "PGauxiliaryFunctions.pl",

           "MathObjects.pl"

            );

TEXT(beginproblem());

$showPartialCorrectAnswers = 1;

Context("Numeric");

$denom= 0;

while ($denom <= 0 ) {

$t1 = random(-4,4,1);

$a=   non_zero_random(-4,4,1);

$a2 = $a*2;

$b=   non_zero_random(-4,4,1);

$gcd = ($a2,$b);

$a3 = $a2/$gcd;

$b3 = $b/$gcd;

$gc8 = $gcd*(-8);

$at3 = $a3*$t1;

$tsq = $t1*$t1;

$bt1 = $b*$t1;

$abt3 = $at3+$b3;

$atsq = $a*$tsq;

$c=   non_zero_random(-4,4,1);

$btc = $bt1 + $c;

$num =$gc8*$abt3;

$den1 = $atsq+$btc;

$den2 = $den1**9;

$gcd = gcd($num,$den2);

$num3 =$num/$gcd;

$den3 =$den2/$gcd;

$ans = $num/$den1**9;

$d =  random(-4,4,1);

$denom = $a*$t1**2 +$b*$t1 +$c;

}

$fp= Compute("-8*$denom**(-9)*(2*$a*$t1 + $b)");

# Present the text.

TEXT(EV2(<<EOT));

Calculate \( f'($t1) \) to 3 significant figures where

\( f'( $t1 ) = \)\{ ans_rule(20) \}

$PAR

Tip: You can enter an answer such as 3.14e-1 for 0.314. $PAR

EOT

ANS($fp->cmp);

SOLUTION(EV3(<<'END_SOLUTION'));

$PAR SOLUTION $PAR

Use the chain rule to find the derivative of composed functions. $BR   

The function of  \( f(t) = ($a t^2 + $b t + $c )^{-8}   \)  is a composed function because the outer function of the negative exponent is "holding" the inner function of a quadratic equation. The chain rule applies the derivative on the outermost function first and then works inward applying the derivative to each successive inner function in a series of products. With each application of the derivative on the outer function the inner contents are left alone. In this way a "chain" of derivative products are formed. The chain rule can be expressed with notation as shown below: $BR

 $BR

Given that `p(x) = f(g(x))`, $BR

$BR

The derivative is applied by the chain rule: $BR

$BR

\( p'(x) = f'(g(x)) \cdot g'(x) \)   $BR

$BR

Apply the derivative to `f(t)`: $BR

$BR

\( f(t) = ($a t^2 + $b t + $c )^{-8}\) $BR

$BR

Apply the prime tics for the chain rule:$BR

$BR

\( f'(t) = \left(($a t^2 + $b t + $c )^{-8}\right)' \left($a t^2 + $b t + $c \right)'\) $BR

$BR

Apply the derivative: $BR

$BR

\( f'(t) = -8($a t^2 + $b t + $c )^{-9} \left($a2 t + $b \right)\) $BR

$BR

Push the quantity of the negative exponent down to the denominator: $BR

$BR

\( \displaystyle{f'(t) = \frac{-8($a2 t + $b) }{($a t^2 + $b t + $c )^{9}} } \) $BR

$BR

Pull any common factors from the numerator: $BR

$BR

\( \displaystyle{f'(t) = \frac{-8 \cdot $gcd($a3 t + $b3) }{($a t^2 + $b t + $c )^{9}} } \) $BR

$BR

Combine factors for the reduced derivative: $BR

$BR

\( \displaystyle{f'(t) = \frac{$gc8($a3 t + $b3) }{($a t^2 + $b t + $c )^{9}}}  \) $BR

$BR

Evaluate the derivative at `t = $t1`: $BR

$BR

\( \displaystyle{f'($t1) = \frac{$gc8($a3 \cdot ($t1) + $b3) }{($a \cdot ($t1)^2 + $b \cdot ($t1) + $c )^{9}}}  \) $BR

$BR

Combine factors and squared values:$BR

$BR

\( \displaystyle{f'($t1) = \frac{$gc8($at3 + $b3) }{($a \cdot $tsq + $bt1 + $c )^{9}}}  \) $BR

$BR

Combine factors and terms:$BR

$BR

\( \displaystyle{f'($t1) = \frac{$gc8($abt3) }{($atsq + $btc )^{9}}}  \) $BR

$BR

Combine factors and terms:$BR

$BR

\( \displaystyle{f'($t1) = \frac{$num }{($den1 )^{9}} =\frac{$num }{$den2} }  \) $BR

$BR

An exact value:

$BR

\( \displaystyle{f'($t1) = \frac{$num3 }{$den3} }  \) $BR

$BR

A decimal approximation:$BR

$BR

\( \displaystyle{f'($t1) = $ans}  \) $BR

$BR

END_SOLUTION

ENDDOCUMENT();        # This should be the last executable line in the problem.