Input your answer here:
\n$\\displaystyle \\frac{dy}{dx}= $ [[0]]
\nInput all numbers as integers not as decimals.
", "marks": 0, "extendBaseMarkingAlgorithm": true, "gaps": [{"showCorrectAnswer": true, "vsetRange": [0, 1], "variableReplacements": [], "answerSimplification": "all,!collectNumbers", "failureRate": 1, "notallowed": {"partialCredit": 0, "showStrings": false, "message": "Input all numbers as integers or as fractions, not as decimals.
", "strings": ["."]}, "answer": "(({( - a)} + ( - (2 * x))-{d}y) / ({b} + (2 * y)+{d}x))", "checkingType": "absdiff", "checkVariableNames": false, "unitTests": [], "type": "jme", "showPreview": true, "expectedVariableNames": [], "scripts": {}, "vsetRangePoints": 5, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "checkingAccuracy": 0.001, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "marks": 2}]}], "name": "Praneetha's copy of Nick's copy of Implicit 2", "variables": {"b": {"definition": "c-1", "name": "b", "templateType": "anything", "group": "Ungrouped variables", "description": ""}, "c": {"definition": "random(2..9 except -a+1)", "name": "c", "templateType": "anything", "group": "Ungrouped variables", "description": ""}, "a": {"definition": "-random(1..9)", "name": "a", "templateType": "anything", "group": "Ungrouped variables", "description": ""}, "d": {"definition": "random(-3..3 except 0)", "name": "d", "templateType": "anything", "group": "Ungrouped variables", "description": ""}}, "metadata": {"description": "\n \t\tImplicit differentiation.
\n \t\tGiven $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\n \t\tAlso find two points on the curve where $x=0$ and find the equation of the tangent at those points.
\n \t\t\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "advice": "
Hint:
\nNote that we regard $y$ as a function of $x$. Hence we have (using the Chain Rule): $\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$. And, using the Product Rule: $\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.
\nNow differentiate both sides of the relation with respect to $x$. Below is a worked solution to the problem, but only look at it if you are struggling.
\na) By differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[( \\simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x-{d}y}\\] and hence on further rearranging:
\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x-{d}y) / ({b} + (2 * y)+{d}x)}\\]
\n\n", "statement": "Given the following relation between $x$ and $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\\]
Find $\\dfrac{dy}{dx}$: