// Numbas version: finer_feedback_settings {"name": "Quotient Rule 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"condition": "", "maxRuns": 100}, "advice": "

These questions use the quotient rule.

The quotient rule is defined as

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

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

Worked example using Part a:

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

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

We can therefore say:

$u=x$

and

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

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

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.

Differentiating with respect to $x$, we get:

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

and

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

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.

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

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

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

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

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

Simplify your answers as much as possible.

", "tags": [], "functions": {}, "name": "Quotient Rule 3", "parts": [{"vsetrangepoints": 5, "scripts": {}, "answersimplification": "all", "checkvariablenames": true, "answer": "({c[1]}*{c[3]}-{c[2]})/(x+{c[3]})^2", "showCorrectAnswer": false, "checkingaccuracy": 0.001, "variableReplacements": [], "type": "jme", "prompt": "

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

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$\\simplify{x^{p[0]}}$$\\simplify{{c[4]}x+{c[5]}}$

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An introduction to using the quotient rule

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