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Given the following relation between $x$ and $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{a}x+{b}y}=\\var{c}\\]
answer the following question.

", "tags": ["calculus", "Calculus", "checked2015", "derivative", "derivative ", "deriving an implicit relation", "differentiate", "differentiate implicitly", "differentiation", "first derivative using implicit differentiation", "implicit differentiation", "implicit relation", "mas1601", "MAS1601"], "functions": {}, "question_groups": [{"name": "", "pickQuestions": 0, "pickingStrategy": "all-ordered", "questions": []}], "variables": {"c": {"group": "Ungrouped variables", "definition": "random(1..9)", "templateType": "anything", "description": "", "name": "c"}, "b": {"group": "Ungrouped variables", "definition": "random(1..9)", "templateType": "anything", "description": "", "name": "b"}, "a": {"group": "Ungrouped variables", "definition": "-random(1..9)", "templateType": "anything", "description": "", "name": "a"}}, "name": "Luis's copy of Implicit differentiation", "ungrouped_variables": ["a", "c", "b"], "variable_groups": [], "type": "question", "preamble": {"css": "", "js": ""}, "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"description": "\n \t\t

Implicit differentiation.

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Given $x^2+y^2+ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

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20/06/2012:

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Added tags.

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Improved display using \\displaystyle where appropriate.

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Changed marks to 2.

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3/07/2012:

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Added tags.

\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "parts": [{"type": "gapfill", "prompt": "\n

Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

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Input your answer here:

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

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On differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) + {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[(\\var{b} + 2y) \\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x}\\] and hence on further rearranging:
\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x) / ({b} + (2 * y))}\\]

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