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Use the product rule to differentiate the function $y=2t^2 \\cos (2t)$.

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$\\displaystyle \\frac{dy}{dt}= $ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"4t*cos(2t)+2t^2*sin(2t)\", \"You're on the right track, but it looks like you have made a couple of mistakes. First, double check that you have correctly differentiated $\\\\cos(\\\\cdot)$. Next, note that while the log tables give the rule for $\\\\cos(\\\\textbf{$x$})$ or $\\\\cos(\\\\textbf{$t$})$, here you have $\\\\cos(\\\\color{red}{\\\\textbf{$2$}}t)$. What extra step do you need to correctly differentiate this?\"],\n [\"4t*cos(2t)-2t^2*sin(2t)\", \"Almost there! Note that while the log tables give the rule for $\\\\cos(\\\\textbf{$x$})$ or $\\\\cos(\\\\textbf{$t$})$, here you have $\\\\cos(\\\\color{red}{\\\\textbf{$2$}}t)$. What extra step do you need to correctly differentiate this?\"],\n [\"4t*cos(2t)+4t^2*sin(2t)\", \"Almost there! Double check that you have correctly differentiated $\\\\cos(\\\\cdot)$ - in particular, note the signs.\"],\n [\"4t*cos(2t)-4t^2*sin(t)\", \"Very close. Remember, the angle of a trig. function never changes when you differentiate!\"],\n [\"4t*cos(2t)+4t^2*sin(t)\", \"You're on the right track, but need to watch a couple of things. First, remember that the angle of a trig. function never changes when you differentiate. Next, double check that you have correctly differentiated $\\\\cos(\\\\cdot)$.\"],\n [\"4t*cos(2t)+2t^2*sin(t)\", \"You're on the right track, but need to watch a couple of things. First, remember that the angle of a trig. function never changes when you differentiate. Next, double check that you have correctly differentiated $\\\\cos(\\\\cdot)$. Finally, note that while the log tables give the rule for $\\\\cos(\\\\textbf{$x$})$ or $\\\\cos(\\\\textbf{$t$})$, here you have $\\\\cos(\\\\color{red}{\\\\textbf{$2$}}t)$. What extra step do you need to correctly differentiate this?\"],\n [\"4t*cos(2t)-2t^2*sin(t)\", \"You're on the right track, but need to watch a couple of things. First, remember that the angle of a trig. function never changes when you differentiate. Also, note that while the log tables give the rule for $\\\\cos(\\\\textbf{$x$})$ or $\\\\cos(\\\\textbf{$t$})$, here you have $\\\\cos(\\\\color{red}{\\\\textbf{$2$}}t)$. What extra step do you need to correctly differentiate this?\"],\n [\"-8t*sin(2t)\", \"The function you need to differentiate is $2t^2 \\\\times \\\\cos(2t)$ i.e. one function of $t$ multiplied by another function of $t$. Therefore, you need the product rule.\"],\n [\"8t*sin(2t)\", \"The function you need to differentiate is $2t^2 \\\\times \\\\cos(2t)$ i.e. one function of $t$ multiplied by another function of $t$. Therefore, you need the product rule.\"],\n [\"4t*sin(2t)\", \"The function you need to differentiate is $2t^2 \\\\times \\\\cos(2t)$ i.e. one function of $t$ multiplied by another function of $t$. Therefore, you need the product rule.\"],\n [\"-4t*sin(2t)\", \"The function you need to differentiate is $2t^2 \\\\times \\\\cos(2t)$ i.e. one function of $t$ multiplied by another function of $t$. Therefore, you need the product rule.\"]\n ]\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))