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We are asked to differentiate:
\n\\[ y=\\frac{e^{\\var{b}x}}{x^{\\var{a}}} \\]
\n\nAt first this function looks like two functions divided rather than multiplied, making it a candidate for the Quotient Rule instead of the Product Rule.
\nHowever, recognising that the function $\\frac{1}{x^{\\var{a}}}$ (the denominator) can be rewritten as $x^{-\\var{a}}$ gives us:
\n\\[ y=e^{\\var{b}x} x^{-\\var{a}}\\]
\nWe are now on much more familiar ground for the Product Rule.
\n\n$u$ is the first function, $v$ is second:
\n\n$\\large u=e^{\\var{b}x} $ $ \\large v=x^{-\\var{a}}$
\n\n
Now, we need to use the approriate techniques to differentiate each of these, for these functions we need your Table of Derivatives:
\n\nThis gives us:
\n$\\large \\frac{du}{dx}=\\var{b}e^{\\var{b}x} $ and $ \\large \\frac{dv}{dx}= \\simplify{ -{a}x^{-{a}-1} }$
\n\nWe now use the formula:
\n$ \\large \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} $
\n\nMake the appropriate substitutions into the formula:
\n\n$ \\large \\frac{dy}{dx}= e^{\\var{b}x} \\times \\simplify{-{a}x^{-{a}-1} } + x^{-\\var{a}} \\times \\var{b}e^{\\var{b}x} $
\n\n\n
Finally, we need to use our basic algebra to simplify this as much as possible. Multiply out brackets where it would simplify and collect like terms:
\n\n$ \\large \\frac{dy}{dx}= -\\frac{\\var{a} e^{\\var{b}x} }{\\simplify{x^{{a}+1}}}+\\frac{\\var{b}e^{\\var{b}x}}{x^{\\var{a}}}$
\n\nNotice that, whenever possible, your final answer should not contain negative indices (powers).
\n\n", "rulesets": {}, "extensions": [], "variables": {"b": {"name": "b", "group": "Ungrouped variables", "definition": "random(2..6)", "description": "
aP
", "templateType": "anything"}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(2..6 except b)", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["b", "a"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "First identify the two functions $u$ and $v$:
\n$u=$[[0]] $v=$[[1]]
\n\n
Now differentiate each one:
\n$ \\large \\frac{du}{dx}= $[[2]] $ \\large\\frac{dv}{dx}= $[[3]]
\n\nThen using:
\n$ \\large \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} $
\n\n
Substitute each component into the formula in the correct place:
\n$ \\large \\frac{dy}{dx}=$[[4]]$ \\large \\times$[[5]]$ \\large + $[[6]]$ \\large \\times$[[7]]
\n\n
Finally tidy this up to give your final answer:
\n$ \\large \\frac{dy}{dx}= $[[8]]
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