Please ensure you are using the correct browser to do this quiz. There is a time limit of 50 minutes to complete this quiz but you may pause the quiz and resume and a later time. The pass mark is 90%. You have unlimited attempts. The quiz is worth 2% of your continuous assessment.
\nN.B. If you pass the quiz do not enter the quiz again as it will reset your score to 0. If you want to check your grade do so through the gradebook link in moodle. Take a screenshot of your pass mark at the end to have as proof that you have passed the quiz and email to yourself just in case.
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\\[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))}\\]
Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\nInput your answer here:
\n$\\displaystyle \\frac{dy}{dx}= $ [[0]]
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\\[\\simplify[all,!collectNumbers]{x^2+y^2+{a}x+{b}y}=\\var{c}\\]
answer the following question.
20/06/2012:
\n \t\tAdded tags.
\n \t\tImproved display using \\displaystyle where appropriate.
\n \t\tChanged marks to 2.
\n \t\t\n \t\t
3/07/2012:
Added tags.
\n \t\t", "description": "\n \t\tImplicit differentiation.
\n \t\tGiven $x^2+y^2+ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
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Find the equation of the tangent line to the curve $f(x) = \\var{a}x^2 - \\var{b}x$ at the point $(\\var{c},\\simplify{{a}{c}{c}-{b}{c}})$.
\n$y = $[[0]]
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\n$y = $ [[0]]
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\n$y = $ [[0]]
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\n$y = $ [[0]]
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\n$\\frac{dy}{dx} = $ [[0]]
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\n$\\frac{dy}{dx} = $[[0]]
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\n$\\frac{dy}{dx} = $[[0]].
\nWhen writing $xe^y$ in your answer be sure to include a * symbol for multiplication i.e. x*e^y.
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\n$\\frac{dy}{dx} = $ [[0]].
\nWhen writing $x\\cos(y)$ in your answer be sure to include a * symbol for multiplication i.e. x*cos(y).
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\n$y'(x) = $ [[0]].
\n$y''(x) = $ [[1]].
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