// Numbas version: finer_feedback_settings {"name": "CF Maths Portfolio - Differentiation 11 - Product Rule (Basic)", "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", "c1"], "name": "CF Maths Portfolio - Differentiation 11 - Product Rule (Basic)", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

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

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

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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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Introduction to using the product rule

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