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{"name": "Chain Rule 02", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Table_of_Derivatives_PyuBTy5.pdf", "/srv/numbas/media/question-resources/Table_of_Derivatives_PyuBTy5.pdf"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Chain Rule 02", "tags": [], "metadata": {"description": "Instructional \"drill\" exercise to emphasize the method.", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "<h4>We use the CHAIN RULE (also called the FUNCTION OF A FUNCTION RULE) when the function that we need to differentiate is actually one function \"nested\" inside another.</h4>\n<p>If $y=f(g(x))$&nbsp; to find $\\frac{dy}{dx}$ , we need to do two things::</p>\n<ol>\n<li>Substitute $u=g(x)$&nbsp; so that $y=f(u)$</li>\n<li>Use the Chain Rule:</li>\n</ol>\n<p>\\[ \\large&nbsp; \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du}&nbsp; \\]</p>\n<p></p>\n<hr/>\n<p></p>", "advice": "<p>We are asked to differentiate:</p>\n<p>\\[&nbsp;y=\\cos{(x^{\\var{bP}})}&nbsp;&nbsp; \\]</p>\n<p></p>\n<p>Recognising that this&nbsp;is a \"<strong>function of another function</strong>\", we need to identify the \"innermost\" of the two functions that are involved and substitute $u$</p>\n<p></p>\n<p style=\"text-align: center;\">Let&nbsp; &nbsp;$u=x^{\\var{bP}}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; then&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $y=cos(u)$</p>\n<p></p>\n<p>Now, we need to use the approriate techniques to differentiate each of these, for&nbsp;the first we need the <strong>Power Rule, </strong>and for the second you can use your<strong> <a href=\"resources/question-resources/Table_of_Derivatives_PyuBTy5.pdf\" target=\"_blank\" rel=\"noopener noreferrer\">Table of Derivatives</a>.</strong></p>\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: left;\">Applying this method gives us:</p>\n<p style=\"text-align: center;\">$\\large \\frac{du}{dx}=\\simplify{{bP}x^{bP2}}$&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; and&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $&nbsp;\\large \\frac{dy}{du}= -sin(u)$</p>\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: left;\">&nbsp;</p>\n<p style=\"text-align: left;\">We now use the Chain Rule formula:</p>\n<p style=\"text-align: center;\">$&nbsp;\\large&nbsp; \\large&nbsp; \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du}&nbsp; &nbsp;$</p>\n<p style=\"text-align: left;\">Make the appropriate substitutions into the formula:</p>\n<p style=\"text-align: left;\"></p>\n<p style=\"text-align: center;\">$ \\large&nbsp; \\frac{dy}{dx}= \\simplify{{bP}x^{bP2}} \\times&nbsp; -sin(u)$</p>\n<p style=\"text-align: left;\"></p>\n<p style=\"text-align: left;\">&nbsp;</p>\n<p style=\"text-align: left;\">Which simplifies to:</p>\n<p style=\"text-align: center;\">$&nbsp;\\large&nbsp; \\frac{dy}{dx}=-\\simplify{ {bP}x^{bP2}}sin(u)&nbsp; $</p>\n<p style=\"text-align: left;\"></p>\n<p style=\"text-align: left;\">Now, finally, we must remember that $u$ was a variable that we introduced and was not part of the original problem.&nbsp;</p>\n<p style=\"text-align: left;\"></p>\n<p style=\"text-align: left;\">Replace $u$ from our original substitution to give the final answer:</p>\n<p style=\"text-align: left;\"></p>\n<p style=\"text-align: center;\">$\\large&nbsp; \\frac{dy}{dx}=-\\simplify{{bP}x^{bP2}} sin(x^{\\var{bP}})$</p>\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: left;\"></p>\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: center;\">&nbsp;</p>\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: center;\">&nbsp;</p>", "rulesets": {}, "extensions": [], "variables": {"bP": {"name": "bP", "group": "Part (b)", "definition": "random(2 .. 5#1)", "description": "<p>Part b) x power</p>", "templateType": "randrange"}, "bP2": {"name": "bP2", "group": "Part (b)", "definition": "bP-1", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "Part (b)", "variables": ["bP", "bP2"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "<h4 style=\"text-align: left;\">Differentiate $&nbsp; &nbsp;y=\\cos{(x^{\\var{bP}})} $</h4>\n<p>First identify the \"innermost\" function, and substitute $u$:</p>\n<p style=\"text-align: center;\">Let&nbsp; &nbsp;$u=$[[0]]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Then&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $y=$[[1]]</p>\n<p style=\"text-align: left;\">Then:</p>\n<p style=\"text-align: center;\">$&nbsp; \\large&nbsp; \\frac{du}{dx}=&nbsp; $[[2]]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; and&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $&nbsp; \\large&nbsp;&nbsp;\\frac{dy}{du}=&nbsp; $[[3]]</p>\n<p style=\"text-align: left;\">Now using:</p>\n<p style=\"text-align: center;\">$\\large&nbsp; \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du}$</p>\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: center;\">$\\large&nbsp; \\frac{dy}{dx}=$[[4]]$\\times$[[5]]</p>\n<p style=\"text-align: left;\">Which simplifies to:</p>\n<p style=\"text-align: center;\">$\\large&nbsp; \\frac{dy}{dx}=$[[6]]</p>\n<p style=\"text-align: left;\"></p>\n<p style=\"text-align: left;\">Remember that $u$ was a variable that we introduced and not part of the original problem.&nbsp;</p>\n<p style=\"text-align: left;\">Replace $u$ from our substitution to give the final answer:</p>\n<p style=\"text-align: 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