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Implicit differentiation question with customised feedback to catch some common errors.

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Find the gradient of the tangent to the curve $\\ln(y^3)+xy=1$ at the point $(1,1)$.

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First find $\\displaystyle \\frac{dy}{dx}$:

\n

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"3/y+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1+3/y)\", \"There are two main errors here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"],\n [\"-y/(x+1/y^3)\", \"Be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/y^3+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1/y^3+1)\", \"There are three things to watch here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$. Finally, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))Now, evaluate the derivative at the point $(1,1)$:

\n

$\\displaystyle\\frac{dy}{dx} \\bigg|_{(1,1)}=$ [[0]]

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