Let \\[\\simplify[std]{f(x) = ({a} * x+{b})/({c}x+{d})}\\]
", "name": "Luis's copy of Quotient rule - differentiate quotient of linear terms", "metadata": {"notes": "\n\t\t1/08/2012:
\n\t\tAdded tags.
\n\t\tChecked calculation. OK.
\n\t\tAdded description.
\n\t\tAll round improvement in display.
\n\t\tAdded forbidden instructions on using decimals.
\n\t\tAdded information on losing 1 mark if use Show steps in part a).
\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Other method. Find $p,\\;q$ such that $\\displaystyle \\frac{ax+b}{cx+d}= p+ \\frac{q}{cx+d}$. Find the derivative of $\\displaystyle \\frac{ax+b}{cx+d}$.
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\\[\\simplify[std]{f(x) = a + b/({c}x+{d})}\\]
Enter a and b as integers or fractions, but not as decimals.
$a=\\;$[[0]]
\n\t\t\t$b=\\;$[[1]]
\n\t\t\tYou can click on Show steps to get some help, but you will lose 1 mark if you do so.
\n\t\t\t", "type": "gapfill", "marks": 0, "showCorrectAnswer": true, "gaps": [{"checkvariablenames": false, "showCorrectAnswer": true, "answer": "{a}/{c}", "vsetrange": [0, 1], "expectedvariablenames": [], "vsetrangepoints": 5, "type": "jme", "scripts": {}, "checkingtype": "absdiff", "notallowed": {"partialCredit": 0, "message": "Input numbers as fractions or integers and not as decimals.
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", "type": "information", "marks": 0, "scripts": {}, "showCorrectAnswer": true}]}, {"prompt": "\n\t\t\tDifferentiate
\\[\\simplify[std]{g(x) = 1/({c}x+{d})}\\]
$\\displaystyle \\frac{dg}{dx}=\\;$[[0]]
\n\t\t\tHence using the first part of the question differentiate \\[\\simplify[std]{f(x) = ({a} * x+{b})/({c}x+{d})}\\]
\n\t\t\t$\\displaystyle \\frac{df}{dx}=\\;$[[1]]
\n\t\t\tInput numbers as fractions or integers and not as decimals.
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\n\tWe have $\\displaystyle \\simplify[std]{{a}x+{b}={a}/{c}*({c}x+{d})+{b}-{a}*{d}/{c}={a}/{c}*({c}x+{d})+{-det}/{c}}$
Hence \\[\\begin{eqnarray*} \\simplify[std]{({a} * x+{b})/({c}x+{d})}&=&\\simplify[std]{({a}/{c}*({c}x+{d})+{-det}/{c})/({c}x+{d})}\\\\ &=&\\simplify[std]{{a}/{c}+({-det}/{c})/({c}x+{d})} \\end{eqnarray*}\\]
Where we have divided out by $\\simplify[std]{{c}x+{d}}$ at the last step.
b)
\n\tWe have \\[\\frac{dg}{dx} = \\simplify[std]{{-c}/({c}x+{d})^2}\\]
using standard rules of differentiation.
Since from a), \\[f(x) = \\simplify[std]{{a}/{c}+({-det}/{c})/({c}x+{d})}\\]
we see that
\\[\\begin{eqnarray*}\\frac{df}{dx} &=&\\simplify[std,!unitPower,!unitDenominator,!zeroFactor,!zeroTerm,!zeroPower]{(-{c})*(({-det}/{c})/({c}x+{d})^2)}\\\\ &=&\\simplify[std]{{det}/({c}x+{d})^2} \\end{eqnarray*}\\]