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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 are asked to differentiate:
\n\\[ \\large y=\\frac{e^{\\var{aP}t}}{t+\\var{aC1}} \\]
\n\nRecognising that the function to differentiate is a quotient, we identify the two functions that are involved.
\n\n$u$ is the numerator, the function \"on top\", $v$ is the denominator, the function \"on the bottom\".
\n\n$u=e^{\\var{aP}t}$ $v=t+\\var{aC1}$
\n\n
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:
\n\nif $ \\frac{d}{dx} [a^x] = a^x \\ln{(a)}$
\n\nIf you don't follow this, use your Table of Derivatives.
\n\nApplying this gives us:
\n$\\large \\frac{du}{dx}=\\simplify{{aP}e^({aP}t)}$ and $\\frac{dv}{dx}=1$
\n\n\n
Make the appropriatesubstitutions into the formula:
\n\n$ \\large \\frac{dy}{dx}= \\frac{(t+\\var{aC1}) \\times (\\simplify{{aP}e^({aP}t)}) - e^{\\var{aP}t} }{(t+\\var{aC1})^2} $
\n\n\n
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:
\n\n\n$ \\large \\frac{dy}{dx}= \\frac{\\simplify{{aP}e^({aP}t)}(t+\\var{aC1}) - (e^{\\var{aP}t}) }{(t+\\var{aC1})^2}$
\n\n", "rulesets": {"std": ["all"]}, "extensions": [], "variables": {"aC1": {"name": "aC1", "group": "Part (a)", "definition": "random(2 .. 6#1)", "description": "
Part a) Constant term for denominator
\nPart a) x power for denominator
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\n$u=$[[0]] $v=$[[1]]
\n\n
Now differentiate each one:
\n$ \\large \\frac{du}{dt}= $[[2]] $ \\large\\frac{dv}{dt}= $[[3]]
\n\n
Then using:
\n$ \\Large \\frac{dy}{dx}= \\frac{v \\frac{du}{dx} - u \\frac{dv}{dx}}{v^{2}} $
\n\nSubstitute each component into the formula in the correct place:
\n\n | $ \\Large \\frac{dy}{dt}=$ | \n
Finally tidy this up to give your final answer:
\n$ \\Large \\frac{dy}{dx}= $ [[9]]
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