\n

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

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\n(Answer in fraction form if necessary)

", "marks": 0, "type": "gapfill", "customName": "", "customMarkingAlgorithm": ""}], "name": "Implicit differentiation 1 (Basic)", "variable_groups": [], "ungrouped_variables": ["a", "b", "c", "d", "constant"], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "Implicit differentiation question with customised feedback to catch some common errors.

"}, "tags": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "advice": "$\\frac{d}{dx}(\\var{a}x+x^2y^2)=\\frac{d}{dx}(\\simplify{{constant}+{b}y})$

\n$\\var{a}+2x.y^2+x^2.2y.\\frac{dy}{dx}=\\var{b}.\\frac{dy}{dx}$

\n$(\\simplify{2x^2y-{b}}) \\frac{dy}{dx}=\\simplify{-{a}-2x*y^2}$

\n$\\frac{dy}{dx}=\\frac{\\simplify{-{a}-2x*y^2}}{\\simplify{2x^2*y-{b}}} =\\simplify{-(2x*y^2+{a})/(2x^2*y-{b})}$

\n$\\ $

\n$\\frac{dy}{dx} \\bigg|_{(\\var{c},\\var{d})}=\\simplify{{a}/{b}}$

\n\n", "statement": "Find the gradient of the curve $\\simplify{{a}x}+x^2y^2=\\simplify{{constant}+{b}y}$ at the point $(\\var{c},\\var{d})$.

\n", "type": "question", "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}, {"name": "JPO AddMath", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2346/"}]}]}], "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}, {"name": "JPO AddMath", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2346/"}]}