// Numbas version: finer_feedback_settings {"name": "Jinhua's copy of Differentiation 13 - Quotient Rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["c", "p"], "name": "Jinhua's copy of Differentiation 13 - Quotient Rule", "tags": [], "advice": "
These questions use the quotient rule.
\nThe quotient rule is defined as
\n$\\frac{dy}{dx}=\\frac{v\\frac{du}{dx}+u\\frac{dv}{dx}}{v^2}$
\nwhen $y=\\frac{u}{v}$
\nWorked example using Part a:
\nThis expression is the result of $x$ divided by ($\\simplify{x+{c[0]}}$).
\nWe can therefore say:
\n$u=x$
\nand
\n$v=\\simplify{x+{c[0]}}$,
\nHence meaning that $y=\\frac{u}{v}$.
\n\nWe already have what $u$ and $v$ equal, so all we have to do is find what $\\frac{du}{dx}$ and $\\frac{dv}{dx}$ are, and then substitute everything into the rule.
\nDifferentiating with respect to $x$, we get:
\n$\\frac{du}{dx}=1$
\nand
\n$\\frac{dv}{dx}=1$.
\nAs there are no powers or coefficients of $x$ that are $>1$, this is a very simple version of the quotient rule, but knowing how to work out this equation formally will make more difficult looking problems just as simple.
\nSubstituting in all the results we've found, we get:
\n$\\frac{dy}{dx}=\\frac{1(\\simplify{x+{c[0]}})+1(x)}{\\simplify{(x+{c[0]})^2}}$
\nWe then simplify, collecting all the terms, to get our final answer of:
\n$\\frac{dy}{dx}=\\simplify{((2x+{c[0]}))/(x+{c[0]})^2}$
", "rulesets": {"std": ["all"]}, "parts": [{"variableReplacements": [], "prompt": "Differentiate the following expressions with respect to $x$ using the quotient rule.
\nSimplify your answers as much as possible.
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", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "contributors": [{"name": "Jinhua Mathias", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/353/"}]}]}], "contributors": [{"name": "Jinhua Mathias", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/353/"}]}