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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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", "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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First identify the two functions  $u$  and  $v$:

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$u=$[[0]]                    $v=$[[1]]

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Now differentiate each one:

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$  \\large \\frac{du}{dt}=   $[[2]]                    $  \\large\\frac{dv}{dt}=   $[[3]]

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Then using:

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

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Substitute each component into the formula in the correct place:

\n\n\n\n\n\n\n\n\n
                                                               $  \\Large \\frac{dy}{dt}=$[[4]]$\\times$ [[5]]$\\large -$ [[6]]$\\times$ [[7]][[8]]
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Finally tidy this up to give your final answer:

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$  \\Large \\frac{dy}{dx}=   $ [[9]]

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