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If $y=f(g(x))$ to find $\\frac{dy}{dx}$ , we need to do two things::
\n\\[ \\large \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du} \\]
\n\nWe are asked to differentiate:
\n\\[ y=(\\var{aCF}x-\\var{C})^{\\var{aP}} \\]
\n\nRecognising that this is a \"function of another function\", we need to identify the \"innermost\" of the two functions that are involved and substitute $u$
\n\nLet $u=\\var{aCF}x-\\var{C}$ then $y=u^{\\var{aP}}$
\n\nNow, we need to use the approriate techniques to differentiate each of these, for both of these we only need the Power Rule:
\n\nApplying this method gives us:
\n$\\large \\frac{du}{dx}=\\var{aCF}$ and $ \\large \\frac{dy}{du}= \\simplify{{aP}u^({aP}-1)}$
\n\n\n
We now use the Chain Rule formula:
\n$ \\large \\large \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du} $
\nMake the appropriate substitutions into the formula:
\n\n$ \\large \\frac{dy}{dx}= \\var{aCF} \\times \\simplify{{aP}u^({aP}-1)} $
\n\n\n
Which simplifies to:
\n$ \\large \\frac{dy}{dx}=\\simplify{ {aCF}*{aP}u^({aP}-1) }$
\n\nNow, finally, we must remember that $u$ was a variable that we introduced and was not part of the original problem.
\n\nReplace $u$ from our original substitution to give the final answer:
\n\n$\\large \\frac{dy}{dx}=\\simplify{ {aCF}*{aP}({aCF}x-{C})^({aP}-1) }$
\n\n\n\n\n\n\n
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Part a) x coefficient
", "templateType": "randrange"}, "C": {"name": "C", "group": "Part (a)", "definition": "random(5..15 except aCF)", "description": "", "templateType": "anything"}, "aP": {"name": "aP", "group": "Part (a)", "definition": "random(10..18 except aCF except C)", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "Part (a)", "variables": ["C", "aP", "aCF"]}], "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": "First identify the \"innermost\" function, and substitute $u$:
\nLet $u=$[[0]] Then $y=$[[1]]
\nThen:
\n$ \\large \\frac{du}{dx}= $[[2]] and $ \\large \\frac{dy}{du}= $[[3]]
\nNow using:
\n$\\large \\frac{dy}{dx}=\\frac{du}{dx} \\times \\frac{dy}{du}$
\n\n$\\large \\frac{dy}{dx}=$[[4]]$\\times$[[5]]
\nWhich simplifies to:
\n$\\large \\frac{dy}{dx}=$[[6]]
\n\nRemember that $u$ was a variable that we introduced and not part of the original problem.
\nReplace $u$ from our substitution to give the final answer:
\n$\\large \\frac{dy}{dx}=$[[7]]
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