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This quiz asks questions on basic techniques of differentation and some introductory applications.
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"}, "statement": "Differentiate the function:
\n\\(f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x^{\\var{b2}}+\\var{c1}\\)
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\n\\(y=ax^n\\,\\,\\,then\\,\\,\\,\\frac{dy}{dx}=nax^{n-1}\\)
\nIn this example
\n\\(f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x^{\\var{b2}}+\\var{c1}\\)
\n\\(\\frac{df}{dx}=\\var{a2}*\\var{a1}x^{\\var{a2}-1}+\\var{b2}*\\var{b1}x^{\\var{b2}-1}\\)
\n\\(\\frac{dy}{dx}=\\simplify{{a2}*{a1}x^{{a2}-1}+{b2}*{b1}x^{{b2}-1}}\\)
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", "licence": "Creative Commons Attribution 4.0 International"}, "advice": "\\(f(x)=\\frac{\\var{a1}}{x^{\\var{a2}}}+\\sqrt[\\var{a3}]{x}\\)
\n\\(f(x)=\\var{a1}x^{-\\var{a2}}+{x}^{\\frac{1}{\\var{a3}}}\\)
\n\\(\\frac{df}{dx}=-\\var{a2}*\\var{a1}x^{-\\var{a2}-1}+\\frac{1}{\\var{a3}}{x}^{\\frac{1}{\\var{a3}}-1}\\)
\n\\(\\frac{df}{dx}=-\\simplify{{a2}*{a1}x^{-{a2}-1}}+\\frac{1}{\\var{a3}}{x}^{\\simplify{{1-{a3}}/{a3}}}\\)
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\n\\(f(x)=\\frac{\\var{a1}}{x^{\\var{a2}}}+\\sqrt[\\var{a3}]{x}\\)
", "type": "question"}, {"name": "Slope of a curve at a point", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "preamble": {"js": "", "css": ""}, "statement": "Calculate the slope of the curve
\n\\(f(x)=\\var{a}x^3-\\var{b}x^2+\\var{c}x+\\var{d}\\)
\nat the point where \\(x=\\var{f}\\).
", "parts": [{"prompt": "Input your answer correct to one decimal place.
\n\\(slope = \\) [[0]]
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\nThe equation for the slope of a curve is found by differentiating the function.
\n\\(\\frac{df}{dx}=3*\\var{a}x^2-2*\\var{b}x+\\var{c}\\)
\nTo find the slope at a particular point we simply insert the x-coordinate value into this equation.
\nSlope = \\(3*\\var{a}*\\var{f}^2-2*\\var{b}*\\var{f}+\\var{c}\\)
\nSlope = \\(\\simplify{3*{a}*{f}^2-2*{b}*{f}+{c}}\\)
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\n\\(x=\\) [[0]]
\nInput the larger of the two \\(x\\) values.
\n\\(x=\\) [[1]]
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", "ungrouped_variables": ["a", "b", "c"], "advice": "\\(f(x)=2x^3-\\simplify{3*({a}+{b})x^2+6*{a}*{b}x}+\\var{c}\\)
\nTo locate a turning point, differentite the function, set equal to zero and solve.
\n\\(f'(x)=6x^2-\\simplify{6*({a}+{b})x+6*{a}*{b}}=0\\)
\nDivide across by 6 to get the quadratic equation
\n\\(x^2-\\simplify{({a}+{b})x+{a}*{b}}=0\\)
\nThis has factors
\n\\((x-\\var{a})(x-\\var{b})=0\\)
\n\\(x-\\var{a}=0\\) or \\(x-\\var{b}=0\\)
\n\\(x=\\var{a}\\) or \\(x=\\var{b}\\)
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\nRecall the chain rule: \\(\\frac{df}{dx}=\\frac{df}{du}.\\frac{du}{dx}\\)
\nlet \\(u=\\var{a2}x^{\\var{a3}}+\\var{a4}\\) then \\(f(x)=u^\\var{a1}\\)
\n\\(\\frac{df}{du}=\\var{a1}u^{\\var{a1}-1}\\) and \\(\\frac{du}{dx}=\\var{a3}*\\var{a2}x^{\\var{a3}-1}\\)
\n\\(\\frac{df}{dx}=\\var{a1}u^{\\simplify{{a1}-1}}.\\simplify{{a2}*{a3}x^{{a3}-1}}\\)
\n\\(\\frac{df}{dx}=\\simplify{{a1}*{a2}*{a3}x^{{a3}-1}}({\\var{a2}x^{\\var{a3}}+\\var{a4}})^{\\simplify{{a1}-1}}\\)
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\n\\(f(x)=({\\var{a2}x^{\\var{a3}}+\\var{a4}})^\\var{a1}\\)
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", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question"}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "functions": {}, "ungrouped_variables": ["a1", "a2", "a3", "a4", "a5"], "tags": [], "advice": "\\(f(x)=(\\var{a1}x^\\var{a2}+\\var{a3})e^{\\var{a4}x+\\var{a5}}\\)
\nRecall the product rule if \\(f(x)=u.v\\) where \\(u\\) and \\(v\\) are both functions of \\(x\\) then
\n\\(\\frac{df}{dx}=v.\\frac{du}{dx}+u.\\frac{dv}{dx}\\)
\nlet \\(u=\\var{a1}x^\\var{a2}+\\var{a3}\\) and \\(v=e^{\\var{a4}x+\\var{a5}}\\)
\n\\(\\frac{du}{dx}=\\var{a2}*\\var{a1}x^{\\var{a2}-1}\\) and \\(\\frac{dv}{dx}=\\var{a4}*e^{\\var{a4}x+\\var{a5}}\\)
\n\\(\\frac{df}{dx}=e^{\\var{a4}x+\\var{a5}}*\\var{a2}*\\var{a1}x^{\\var{a2}-1}+(\\var{a1}x^\\var{a2}+\\var{a3})*\\var{a4}*e^{\\var{a4}x+\\var{a5}}\\)
\n\\(\\frac{df}{dx}=\\simplify{e^({a4}x+{a5})*{a1}*{a2}x^{{a2}-1}+({a1}x^{a2}+{a3})*{a4}*e^({a4}x+{a5})}\\)
\n\\(\\frac{df}{dx}=(\\simplify{{a1}*{a4}x^{a2}+{a1}*{a2}x^{{a2}-1}+{a3}*{a4}})\\simplify{e^({a4}x+{a5})}\\)
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\n\\(f(x)=(\\var{a1}x^\\var{a2}+\\var{a3})e^{\\var{a4}x+\\var{a5}}\\)
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", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question"}, {"name": "Quotient rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "functions": {}, "ungrouped_variables": ["a", "b", "c", "f", "d"], "tags": [], "preamble": {"css": "", "js": ""}, "advice": "\\(f(x)=\\frac{\\var{a}x^{\\var{b}}+\\var{f}}{\\var{c}cos(\\var{d}x)}\\)
\nRecall the quotient rule: if \\(y=\\frac{u}{v}\\) where \\(u\\) and \\(v\\) are both functions of \\(x\\) then
\n\\(\\frac{dy}{dx}=\\frac{v\\frac{du}{dx}-u\\frac{dv}{dx}}{v^2}\\)
\nLet \\(u=\\var{a}x^{\\var{b}}+\\var{f}\\) and \\(v=\\var{c}cos(\\var{d}x)\\)
\nthen \\(\\frac{du}{dx}=\\var{b}*\\var{a}x^{\\var{b}-1}\\) and \\(\\frac{dv}{dx}=-\\var{d}*\\var{c}sin(\\var{d}x)\\)
\nPutting these results together as shown in the rule gives:
\n\\(\\frac{df}{dx}=\\frac{(\\var{c}cos(\\var{d}x))*\\var{b}*\\var{a}x^{\\var{b}-1}-(\\var{a}x^{\\var{b}}+\\var{f})*(-\\var{d}*\\var{c}sin(\\var{d}x))}{(\\var{c}cos(\\var{d}x))^2}\\)
\n\\(\\frac{df}{dx}=\\frac{\\simplify{({c}*cos({d}x))*{b}*{a}x^{{b}-1}+({a}x^{{b}}+{f})*({c}*{d}*sin({d}x))}}{(\\var{c}*cos(\\var{d}x))^2}\\)
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\n\\(f(x)=\\frac{\\var{a}x^{\\var{b}}+\\var{f}}{\\var{c}cos(\\var{d}x)}\\)
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"}, "statement": "A missile is launched straight up in the air. The height of the missile, \\(h\\) metres, above the ground \\(t\\) seconds after the launch button is pressed is given by:
\n\\(h=\\var{a}t-4.9t^2\\)
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\nRecall that speed is the rate of change of position with respect to time i.e. \\(v=\\frac{dh}{dt}\\)
\n\\(v=\\frac{dh}{dt}=\\var{a}-2*4.9t\\)
\nwhen \\(t=\\var{b}\\)
\n\\(v=\\var{a}-2*4.9*\\var{b}\\)
\n\\(v=\\simplify{{a}-9.8*{b}}m/s\\)
\n\nThe missile will reach its maximum height when its speed = 0. i.e. \\(v=\\frac{dh}{dt}=\\var{a}-2*4.9t=0\\)
\n\\(\\var{a}=9.8t\\)
\n\\(t=\\var{a}/9.8\\)
\nThe maximum height reached will occur when \\(t=\\simplify{{a}/9.8}\\)
\n\\(h=\\var{a}*\\left(\\simplify{{a}/9.8}\\right)-4.9*\\left(\\simplify{{a}/9.8}\\right)^2\\)
\n\\(h=\\simplify{{a}^2/19.6}\\)
\n\\(h=\\simplify{{{a}/{19.6}^0.5}^2}\\)
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\n\\(v = \\) [[0]]m/s
\nWhat is the maximum height achieved by this missile? Give your answer correct to one decimal place.
\n\\(h = \\) [[1]]m
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