// Numbas version: exam_results_page_options {"name": "Cf Maths In class test two mock paper question 7 (i)", "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": "Cf Maths In class test two mock paper question 7 (i)", "tags": [], "advice": "

These questions use the quotient rule.

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The quotient rule is defined as

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$\\frac{dy}{dx}=\\frac{v\\frac{du}{dx}+u\\frac{dv}{dx}}{v^2}$

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when $y=\\frac{u}{v}$

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Worked example for:

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$x$$\\simplify{x+{c[0]}}$

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This expression is the result of $x$ divided by ($\\simplify{x+{c[0]}}$).

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We can therefore say:

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$u=x$

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and

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$v=\\simplify{x+{c[0]}}$,

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Hence meaning that $y=\\frac{u}{v}$.

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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 quotient 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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$\\frac{dy}{dx}=\\frac{1(\\simplify{x+{c[0]}})+1(x)}{\\simplify{(x+{c[0]})^2}}$

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

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$\\frac{dy}{dx}=\\simplify{((2x+{c[0]}))/(x+{c[0]})^2}$

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$\\simplify{{c[1]}x+{c[2]}}$$\\simplify{{c[3]}+x^{p[1]}}$

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

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Simplify your answers as much as possible.

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An introduction to using the quotient rule

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