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Introduction to using the product rule

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$\\simplify{(x+{c[0]})(x+{c1})}$

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$\\simplify{({c[2]}x+{c[3]})({c[4]}x+{c[5]})}$

", "answer": "({c[2]}{c[4]}*2x)+({c[3]}{c[4]}+{c[2]}{c[5]})", "type": "jme", "checkingaccuracy": 0.001, "vsetrange": [0, 1], "checkingtype": "absdiff", "showpreview": true}, {"checkvariablenames": false, "marks": "2", "scripts": {}, "vsetrangepoints": 5, "variableReplacements": [], "showCorrectAnswer": true, "expectedvariablenames": ["x"], "variableReplacementStrategy": "originalfirst", "answersimplification": "all", "prompt": "

$\\simplify{(x^2+{c[6]})({c[7]}x+{c[8]})}$

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$\\simplify{(x^2+{c[9]})(x^3+{c[10]})}$

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These questions use the product rule.

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The product rule is defined as

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$\\frac{dy}{dx}=v\\frac{du}{dx}+u\\frac{dv}{dx}$

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when $y=uv$

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Worked example using Part a:

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$y=\\simplify{(x+{c[0]})(x+{c1})}$

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This expression is the result of two products; $(\\simplify{x+{c[0]}})$ and $(\\simplify{x+{c1}})$.

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We can therefore say:

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$u=(\\simplify{x+{c[0]}})$

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and

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$v=(\\simplify{x+{c1}})$,

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Hence meaning that $y=uv$.

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We 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.

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Differentiating with respect to $x$, we get:

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$\\frac{du}{dx}=1$

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and

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$\\frac{dv}{dx}=1$.

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As 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.

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Substituting in all the results we've found, we get:

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$\\frac{dy}{dx}=1(\\simplify{x+{c1}})+1(\\simplify{x+{c[0]}})$

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We then simplify, collecting all the terms, to get our final answer of:

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$\\frac{dy}{dx}=\\simplify{2x+{c[0]}+{c1}}$

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Differentiate the following expressions with respect to $x$ using the product rule.

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Simplify your answers as much as possible.

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