// Numbas version: finer_feedback_settings
{"name": "Differentiation 12 - Product Rule (with Chain Rule)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variables": {"p": {"templateType": "anything", "name": "p", "description": "", "definition": "repeat(random(2..6),10)", "group": "Ungrouped variables"}, "c": {"templateType": "anything", "name": "c", "description": "", "definition": "repeat(random(-9..9 except 0),13)", "group": "Ungrouped variables"}}, "statement": "<p><strong>Differentiate the following expressions with respect to $x$ using the both product rule and the chain rule.</strong></p>\n<p><strong></strong>Simplify your answers as much as possible.</p>", "name": "Differentiation 12 - Product Rule (with Chain Rule)", "parts": [{"vsetRangePoints": 5, "customMarkingAlgorithm": "", "showPreview": true, "extendBaseMarkingAlgorithm": true, "customName": "", "variableReplacementStrategy": "originalfirst", "checkingType": "absdiff", "failureRate": 1, "valuegenerators": [{"name": "x", "value": ""}], "showFeedbackIcon": true, "prompt": "<p>$\\simplify{{c[1]}x^3({c[2]}x+{c[3]})^{p[1]}}$</p>", "marks": "2", "checkVariableNames": false, "answerSimplification": "all", "vsetRange": [0, 1], "checkingAccuracy": 0.001, "answer": "3{c[1]}x^2*({c[2]}x+{c[3]})^{p[1]}+{p[1]}{c[2]}{c[1]}x^3*({c[2]}x+{c[3]})^({p[1]}-1)", "showCorrectAnswer": true, "scripts": {}, "type": "jme", "unitTests": [], "useCustomName": false, "variableReplacements": []}, {"vsetRangePoints": 5, "customMarkingAlgorithm": "", "showPreview": true, "extendBaseMarkingAlgorithm": true, "customName": "", "variableReplacementStrategy": "originalfirst", "checkingType": "absdiff", "failureRate": 1, "valuegenerators": [{"name": "x", "value": ""}], "showFeedbackIcon": true, "prompt": "<p>$\\simplify{{c[4]}x^{p[2]}({c[5]}x^2+{c[6]})^{p[3]}}$</p>", "marks": "2", "checkVariableNames": false, "answerSimplification": "all", "vsetRange": [0, 1], "checkingAccuracy": 0.001, "answer": "{p[2]}{c[4]}x^({p[2]}-1)*({c[5]}x^2+{c[6]})^{p[3]}+2{p[3]}{c[4]}{c[5]}x^({p[2]}+1)*({c[5]}x^2+{c[6]})^({p[3]}-1)", "showCorrectAnswer": true, "scripts": {}, "type": "jme", "unitTests": [], "useCustomName": false, "variableReplacements": []}], "functions": {}, "rulesets": {}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "tags": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "extensions": [], "ungrouped_variables": ["c", "p"], "advice": "<p>Patience and careful substitution is all that is necessary to differentiate these types of functions.</p>\n<p>Take the function:&nbsp;&nbsp;&nbsp; $y=4x^2(5x^3+2)^4$</p>\n<p>First label the two separate parts which would be differentiated with the <strong>product rule</strong>, namely:</p>\n<p>$u=4x^2\\;\\;\\;\\text{and}\\;\\;\\;v=(5x^3+2)^4$</p>\n<p>We also know that, by the product rule:</p>\n<p>$\\frac{dy}{dx}=u\\frac{dv}{dx}+v\\frac{du}{dx}$</p>\n<p>Looking at the substitution for $u$, the derivate can be easily found with the power rule:</p>\n<p>$\\frac{du}{dx}=(4\\times2)x^{2-1}=8x^1=8x$</p>\n<p>So far, from&nbsp;the product rule,&nbsp;we have:</p>\n<p>$\\frac{dy}{dx}=(4x^2)\\frac{dv}{dx}+(5x^3+2)^4(8x)$</p>\n<p></p>\n<p>Looking at the expression for $v=(5x^3+2)^4$ we can see that finding&nbsp;$\\frac{dv}{dx}$, however, requires the <strong>chain rule</strong>.</p>\n<p>Our usual&nbsp;subsitution letter $u$ has already been assigned an expression in the product rule part. This time, we can use $t$.</p>\n<p>By the chain rule:</p>\n<p>$\\frac{dv}{dx}=\\frac{dv}{dt}\\times\\frac{dt}{dx}$</p>\n<p>Let $t=5x^3+2$</p>\n<p>This means that: $v=t^4\\;\\;\\;\\therefore\\;\\;\\;\\frac{dv}{dt}=4t^3$</p>\n<p>Also, $t=5x^3+2\\;\\;\\;\\therefore\\;\\;\\;\\frac{dt}{dx}=15x^2$</p>\n<p>It follows that: $\\frac{dv}{dx}=\\frac{dv}{dt}\\times\\frac{dt}{dx}=4t^3\\times15x^2=60x^2t^3$</p>\n<p>Subsituting back the original expression for $t$:</p>\n<p>$\\frac{dv}{dx}=60x^2(5x^3+2)^3$</p>\n<p>We can now input our expression for $\\frac{dv}{dx}$ into the original product rule section.</p>\n<p></p>\n<p>$\\frac{dy}{dx}=(4x^2)\\frac{dv}{dx}+(5x^3+2)^4(8x)\\;\\;\\;\\therefore\\;\\;\\;\\frac{dy}{dx}=(4x^2)(60x^2(5x^3+2)^3)+(5x^3+2)^4(8x)$</p>\n<p>This is simplified further:</p>\n<p>$\\frac{dy}{dx}=240x^4(5x^3+2)^3+8x(5x^3+2)^4=8x(5x^3+2)^3\\left(30x^3+(5x^3+2)\\right)$</p>\n<p></p>\n<p>Finally:</p>\n<p>$\\frac{dy}{dx}=8x(5x^3+2)^3(35x^3+2)$</p>\n<p></p>\n<p></p>", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "<p>Using the chain rule within product rule problems.</p>"}, "type": "question", "contributors": [{"name": "", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/514/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/514/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}