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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{\\var{aCF}e^(x)+\\var{aC1}}{\\var{aC2}-\\var{aCF2}e^(x)}\\]
\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$u=\\simplify{{aCF}e^(x)+{aC1}}$ $v=\\simplify{{aC2}-{aCF2}e^(x)}$
\n\n
Now, we need to use the approriate techniques to differentiate each of these, for these functions we need the Exponential Rule:
\n\nif $ \\frac{d}{dx} [a^x] = a^x \\ln{(a)}$
\n\nIf you don't follow this, use your Table of Derivatives that gives you: $e^{kx}$ differentiates to $k e^{kx}$
\n\nApplying this gives us:
\n$\\large \\frac{du}{dx}=\\simplify{ {aCF}e^(x) }$ and $\\frac{dv}{dx}=\\simplify{ -{aCF2}e^(x) }$
\n\n\n
Make the appropriatesubstitutions into the formula:
\n\n$ \\large \\frac{dy}{dx}= \\frac{(\\var{aC2}-\\var{aCF2}e^(x)) \\times (\\var{aCF}e^(x)) - (\\var{aCF}e^(x)+\\var{aC1}) \\times (-\\var{aCF2}e^(x)) }{(\\var{aC2}-\\var{aCF2}e^(x))^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}= \\simplify{ ( {aC2}*{aCF}e^(x)+ {aC1}*{aCF2}e^(x))/(({aC2}-{aCF2}e^(x))^2) }$
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Part a) Constant term for numerator
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\n$u=$[[0]] $v=$[[1]]
\n\n
Now differentiate each one:
\n$ \\large \\frac{du}{dx}= $[[2]] $ \\large\\frac{dv}{dx}= $[[3]]
\n\n
Then using:
\n$ \\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}{dx}=$ | \n
Finally tidy this up to give your final answer:
\n$ \\Large \\frac{dy}{dx}= $ [[9]]
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