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Introduction to using the product rule
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\nSimplify your answers as much as possible.
", "parts": [{"showpreview": true, "variableReplacementStrategy": "originalfirst", "answersimplification": "all", "vsetrangepoints": 5, "vsetrange": [0, 1], "type": "jme", "checkingtype": "absdiff", "checkingaccuracy": 0.001, "scripts": {}, "prompt": "$\\simplify{(x+{c[0]})(x+{c1})}$
", "variableReplacements": [], "answer": "2x+{c1}+{c[0]}", "marks": "2", "showCorrectAnswer": true, "expectedvariablenames": ["x"], "checkvariablenames": false}, {"showpreview": true, "variableReplacementStrategy": "originalfirst", "answersimplification": "all", "vsetrangepoints": 5, "vsetrange": [0, 1], "type": "jme", "checkingtype": "absdiff", "checkingaccuracy": 0.001, "scripts": {}, "prompt": "$\\simplify{({c[2]}x+{c[3]})({c[4]}x+{c[5]})}$
", "variableReplacements": [], "answer": "({c[2]}{c[4]}*2x)+({c[3]}{c[4]}+{c[2]}{c[5]})", "marks": "2", "showCorrectAnswer": true, "expectedvariablenames": ["x"], "checkvariablenames": false}, {"showpreview": true, "variableReplacementStrategy": "originalfirst", "answersimplification": "all", "vsetrangepoints": 5, "vsetrange": [0, 1], "type": "jme", "checkingtype": "absdiff", "checkingaccuracy": 0.001, "scripts": {}, "prompt": "$\\simplify{(x^2+{c[6]})({c[7]}x+{c[8]})}$
", "variableReplacements": [], "answer": "({c[7]}*3x^2)+(2{c[8]}x)+({c[7]}{c[6]})", "marks": "2", "showCorrectAnswer": true, "expectedvariablenames": ["x"], "checkvariablenames": false}, {"showpreview": true, "variableReplacementStrategy": "originalfirst", "answersimplification": "all", "vsetrangepoints": 5, "vsetrange": [0, 1], "type": "jme", "checkingtype": "absdiff", "checkingaccuracy": 0.001, "scripts": {}, "prompt": "$\\simplify{(x^2+{c[9]})(x^3+{c[10]})}$
", "variableReplacements": [], "answer": "5x^4+3{c[9]}x^2+2{c[10]}x", "marks": "2", "showCorrectAnswer": true, "expectedvariablenames": ["x"], "checkvariablenames": true}], "advice": "These questions use the product rule.
\nThe product rule is defined as
\n$\\frac{dy}{dx}=v\\frac{du}{dx}+u\\frac{dv}{dx}$
\nwhen $y=uv$
\nWorked example using Part a:
\n$y=\\simplify{(x+{c[0]})(x+{c1})}$
\nThis expression is the result of two products; $(\\simplify{x+{c[0]}})$ and $(\\simplify{x+{c1}})$.
\nWe can therefore say:
\n$u=(\\simplify{x+{c[0]}})$
\nand
\n$v=(\\simplify{x+{c1}})$,
\nHence meaning that $y=uv$.
\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 product 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$\\frac{dy}{dx}=1(\\simplify{x+{c1}})+1(\\simplify{x+{c[0]}})$
\nWe then simplify, collecting all the terms, to get our final answer of:
\n$\\frac{dy}{dx}=\\simplify{2x+{c[0]}+{c1}}$
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