// Numbas version: finer_feedback_settings {"name": "test 1", "metadata": {"description": "
This is the test description. Placeholder text can be changed per test.
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", "licence": "All rights reserved"}, "statement": "Let $f(x) = \\simplify{ {a} x^{n} }$.
", "advice": "$f(x)= \\simplify{ x^{n} }$.
\n\nBy power rule:
\n\n$f'(x)= \\simplify{ {n}x^{n-1} }$.
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", "useAlternativeFeedback": true, "answer": "{(n-1) * a} * x^{n-1}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{n * a} * x^{n-1}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "valuegenerators": [{"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "What is $\\int f(x)\\,\\mathrm{d}x$? Use $C$ for the constant of integration.
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\nInput all numbers as fractions or integers. Also do not include brackets in your answers.
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\nThe solution is given by:
\n
$\\simplify[std]{{e6*i}}(\\simplify[std]{{a}})=\\simplify[std]{{a*e6*i}}$
b)
$\\simplify[std]{{a}*{z4}={a*z4}}$
\n
c)
\\[ \\begin{eqnarray*} \\simplify[std,!otherNumbers]{{a}*({a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3)}&=&\\simplify[std]{{a}*{a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3}}\\\\ &=&\\simplify[std]{{a*(a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3)}} \\end{eqnarray*} \\]
d)
This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,
\nand then to multiply the result of that calculation by the third set of parentheses.
\nSo we obtain:
\\[ \\begin{eqnarray*} (\\var{a})(\\var{z1})(\\var{z3})&=&((\\var{a})(\\var{z1}))(\\var{z3})\\\\ &=&(\\var{a*(z1)})(\\var{z3})\\\\ &=&\\var{a*(z1)*(z3)} \\end{eqnarray*} \\]
Differentiate $\\displaystyle (ax^m+b)^{n}$.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the chain rule.
", "advice": "$\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$
\nThe chain rule says that if $f(x)=g(h(x))$ then
\n\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
\nOne way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
\nThen we use the chain rule in the form:
\n\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u}\\]
\nOnce you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
\nFor this example, we let $u=\\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\\simplify[std]{u^{n}}$.
\nThis gives
\n\\begin{align}
\\frac{\\mathrm{d}u}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m -1}} \\\\[1em]
\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u} &= \\simplify[std]{{n}u^{n-1}}
\\end{align}
Hence on substituting into the chain rule above we get:
\n\\begin{align}
\\frac{\\mathrm{d}f}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m-1} * ({n}*u^{n-1})} \\\\
&= \\simplify[std]{{m*a*n}x^{m-1}u^{n-1}} \\\\
&= \\simplify[std]{{m*a*n}x^{m-1}({a}*x^{m}+{b})^{n-1}}
\\end{align}
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.
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\n$\\displaystyle \\frac{\\mathrm{d}f}{\\mathrm{d}x}=$ [[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
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\n\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
\nOne way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
\nThen we use the chain rule in the form:
\n\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f}{\\mathrm{d}u}\\]
\nOnce you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
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