// Numbas version: exam_results_page_options {"name": "Quotient Rule 05 (non-scaffolded)", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Table_of_Derivatives_W9at6dx.pdf", "/srv/numbas/media/question-resources/Table_of_Derivatives_W9at6dx.pdf"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Quotient Rule 05 (non-scaffolded)", "tags": [], "metadata": {"description": "

Instructional \"drill\" exercise to emphasize the method.

Thanks to Christian for his method for use of gaps in fractions.

", "licence": "All rights reserved"}, "statement": "

We use the QUOTIENT RULE when the function that we need to differentiate is actually two functions divided:

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If  $\\large y=\\frac{u}{v}$  then:

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\\[   \\frac{dy}{dx}= \\frac{v \\frac{du}{dx} - u \\frac{dv}{dx}}{v^{2}} \\]

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Now it is time to get out your paper and pencil and try similar questions without the help. Carry out the same steps, lay it out the same way but you only need to input the final answer.

", "advice": "

We are asked to differentiate:

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\\[ \\large y=\\frac{e^{\\var{aP}t}}{t+\\var{aC1}} \\]

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Recognising that the function to differentiate is a quotient, we identify the two functions that are involved.

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$u$ is the numerator, the function \"on top\", $v$ is the denominator, the function \"on the bottom\".

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$u=e^{\\var{aP}t}$                    $v=t+\\var{aC1}$

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Now, we need to use the approriate techniques to differentiate each of these, for $u$ we need the Exponential Rule and for $v$ we need the Power Rule:

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if  $  \\frac{d}{dx} [a^x] = a^x \\ln{(a)}$

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If you don't follow this, use your Table of Derivatives.

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Applying this gives us:

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$\\large \\frac{du}{dx}=\\simplify{{aP}e^({aP}t)}$          and          $\\frac{dv}{dx}=1$

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Make the appropriatesubstitutions into the formula:

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$ \\large  \\frac{dy}{dx}=  \\frac{(t+\\var{aC1}) \\times (\\simplify{{aP}e^({aP}t)}) - e^{\\var{aP}t} }{(t+\\var{aC1})^2}  $

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Finally, we need to use our basic algebra terms to simplify this as much as possible. Multiply out brackets where it would simplify and collect like terms:

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$ \\large \\frac{dy}{dx}= \\frac{\\simplify{{aP}e^({aP}t)}(t+\\var{aC1}) - (e^{\\var{aP}t}) }{(t+\\var{aC1})^2}$ 

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Part a) Constant term for denominator

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", "templateType": "randrange"}, "aP": {"name": "aP", "group": "Part (a)", "definition": "random(2..5 except aC1)", "description": "

Part a) x power for denominator

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Differentiate  $ \\large y=\\frac{e^{\\var{aP}t}}{t+\\var{aC1}}$

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$  \\Large \\frac{dy}{dx}=   $[[0]]

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