// Numbas version: finer_feedback_settings {"name": "Differentiation: Quotient Rule 4", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Differentiation: Quotient Rule 4", "tags": [], "metadata": {"description": "
Calculating the derivative of a function of the form $\\frac{e^{kx}}{ax^2+bx+c}$ using the quotient rule.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Find the derivative of \\[ \\simplify{y=e^({d}x)/({a}x^2+{b}x+{c})}. \\]
", "advice": "If we have a function of the form $y=\\tfrac{u(x)}{v(x)}$, to calculate its derivative we need to use the quotient rule:
\n\\[ \\dfrac{dy}{dx} = \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2}\\,.\\]
\nThis can be split up into steps:
\nFollowing this process, we must first identify $u(x)$ and $v(x)$.
\nAs \\[ \\simplify{y=e^({d}x)/({a}x^2+{b}x+{c})}, \\]
\nlet \\[ u(x) = \\simplify{e^({d}x)} \\quad \\text{and} \\quad v(x)=\\simplify{{a}x^2+{b}x+{c}}.\\]
\nNext, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:
\n\\[ \\dfrac{du}{dx} = \\simplify{{d}e^({d}x)}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{2*a}x+{b}}.\\]
\nSubstituting these results into the quotient rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:
\n\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2} \\\\ \\\\&\\,=\\dfrac{(\\simplify{{a}x^2+{b}x+{c}}) \\times\\simplify{{d}e^({d}x)} - \\simplify{e^({d}x)} \\times (\\simplify{{2*a}x+{b}})}{\\simplify{({a}x^2+{b}x+{c})^2}}. \\end{split}\\]
\n\n{advice}
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\\n\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx}&\\\\,=\\\\dfrac{\\\\simplify{{d}e^({d}x)}(\\\\simplify{{a}x^2+{b}x+{c}}) - \\\\simplify{e^({d}x)}(\\\\simplify{{2*a}x+{b}})}{\\\\simplify{({a}x^2+{b}x+{c})^2}} \\\\\\\\ \\\\\\\\ &\\\\,=\\\\dfrac{\\\\simplify[all,!collectNumbers,!cancelTerms]{e^({d}x)({d*a}x^2+{d*b}x+{d*c}-{2*a}x-{b})}}{\\\\simplify{({a}x^2+{b}x+{c})^2}} \\\\\\\\ \\\\\\\\&\\\\,=\\\\dfrac{\\\\simplify{e^({d}x)({d*a}x^2+{d*b}x+{d*c}-{2*a}x-{b})}}{\\\\simplify{({a}x^2+{b}x+{c})^2}}.\\\\end{split} \\\\]
\\n\"", "description": "", "templateType": "long string", "can_override": false}, "simplify": {"name": "simplify", "group": "Ungrouped variables", "definition": "\"Simplifying,
\\n\\\\[ \\\\begin{split} \\\\dfrac{dy}{dx}&\\\\,=\\\\dfrac{\\\\simplify{{d}e^({d}x)}(\\\\simplify{{a}x^2+{b}x+{c}}) - \\\\simplify{e^({d}x)}(\\\\simplify{{2*a}x+{b}})}{\\\\simplify{({a}x^2+{b}x+{c})^2}} \\\\\\\\ \\\\\\\\ &\\\\,=\\\\dfrac{\\\\simplify[all,!collectNumbers,!cancelTerms]{e^({d}x)({d*a}x^2+{d*b}x+{d*c}-{2*a}x-{b})}}{\\\\simplify{({a}x^2+{b}x+{c})^2}} \\\\\\\\ \\\\\\\\&\\\\,=\\\\dfrac{\\\\simplify{e^({d}x)({d*a}x^2+{d*b}x+{d*c}-{2*a}x-{b})}}{\\\\simplify{({a}x^2+{b}x+{c})^2}} \\\\\\\\ \\\\\\\\&\\\\,=\\\\dfrac{\\\\simplify{{simp}e^({d}x)({d*a/simp}x^2+{d*b/simp}x+{d*c/simp}-{2*a/simp}x-{b/simp})}}{\\\\simplify{({a}x^2+{b}x+{c})^2}}.\\\\end{split} \\\\]
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