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This quiz asks questions on basic techniques of differentation and some introductory applications.

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Power rule

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Differentiate the function:

\n

\\(f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x^{\\var{b2}}+\\var{c1}\\)

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\\(\\frac{df}{dx}=\\) [[0]]

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Apply the rule:

\n

\\(y=ax^n\\,\\,\\,then\\,\\,\\,\\frac{dy}{dx}=nax^{n-1}\\)

\n

In 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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\\(\\frac{df}{dx}=\\) [[0]]

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Power rule

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\\(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}}}\\)

\n

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Differentiate the function:

\n

\\(f(x)=\\frac{\\var{a1}}{x^{\\var{a2}}}+\\sqrt[\\var{a3}]{x}\\)

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Calculate the slope of the curve

\n

\\(f(x)=\\var{a}x^3-\\var{b}x^2+\\var{c}x+\\var{d}\\)

\n

at the point where \\(x=\\var{f}\\). 

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Input your answer correct to one decimal place.

\n

\\(slope = \\) [[0]]

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Slope of a curve at a point

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\\(f(x)=\\var{a}x^3-\\var{b}x^2+\\var{c}x+\\var{d}\\)

\n

The 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}\\)

\n

To find the slope at a particular point we simply insert the x-coordinate value into this equation.

\n

Slope = \\(3*\\var{a}*\\var{f}^2-2*\\var{b}*\\var{f}+\\var{c}\\)

\n

Slope = \\(\\simplify{3*{a}*{f}^2-2*{b}*{f}+{c}}\\)

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\\(f(x)=\\var{a1}sin(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

\n

Recall the chain rule:   \\(\\frac{df}{dx}=\\frac{df}{du}.\\frac{du}{dx}\\)

\n

let \\(u=\\var{a2}x^{\\var{a3}}+\\var{a4}\\)    then   \\(f(x)=\\var{a1}sin(u)\\)

\n

\\(\\frac{df}{du}=\\var{a1}cos(u)\\)  and  \\(\\frac{du}{dx}=\\var{a3}*\\var{a2}x^{\\var{a3}-1}\\)

\n

\\(\\frac{df}{dx}=\\var{a1}cos(u).\\simplify{{a2}*{a3}x^{{a3}-1}}\\)

\n

\\(\\frac{df}{dx}=\\simplify{{a1}*{a2}*{a3}x^{{a3}-1}}cos(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

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\\(\\frac{df}{dx}=\\) [[0]]

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Differentiate the function:

\n

\\(f(x)=\\var{a1}sin(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

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Chain rule

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\\(f(x)=\\var{a1}ln(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

\n

Recall the chain rule:   \\(\\frac{df}{dx}=\\frac{df}{du}.\\frac{du}{dx}\\)

\n

let \\(u=\\var{a2}x^{\\var{a3}}+\\var{a4}\\)    then   \\(f(x)=\\var{a1}ln(u)\\)

\n

\\(\\frac{df}{du}=\\frac{\\var{a1}}{u}\\)  and  \\(\\frac{du}{dx}=\\var{a3}*\\var{a2}x^{\\var{a3}-1}\\)

\n

\\(\\frac{df}{dx}=\\frac{\\var{a1}}{u}.\\simplify{{a2}*{a3}x^{{a3}-1}}\\)

\n

\\(\\frac{df}{dx}=\\frac{\\simplify{{a1}*{a2}*{a3}x^{{a3}-1}}}{\\var{a2}x^{\\var{a3}}+\\var{a4}}\\)

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\\(\\frac{df}{dx}=\\) [[0]]

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Differentiate the function:

\n

\\(f(x)=\\var{a1}ln(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

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Chain rule

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\\(f(x)=({\\var{a2}x^{\\var{a3}}+\\var{a4}})^\\var{a1}\\)

\n

Recall the chain rule:   \\(\\frac{df}{dx}=\\frac{df}{du}.\\frac{du}{dx}\\)

\n

let \\(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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\\(\\frac{df}{dx}=\\)[[0]]

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Differentiate the function:

\n

\\(f(x)=({\\var{a2}x^{\\var{a3}}+\\var{a4}})^\\var{a1}\\)

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Chain rule

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\\(f(x)=(\\var{a1}x^\\var{a2}+\\var{a3})e^{\\var{a4}x+\\var{a5}}\\)

\n

Recall 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}\\)

\n

let \\(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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\\(\\frac{df}{dx}=\\)[[0]]

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Differentiate the function

\n

\\(f(x)=(\\var{a1}x^\\var{a2}+\\var{a3})e^{\\var{a4}x+\\var{a5}}\\)

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Product rule

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\\(f(x)=\\frac{\\var{a}x^{\\var{b}}+\\var{f}}{\\var{c}cos(\\var{d}x)}\\)

\n

Recall 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}\\)

\n

Let   \\(u=\\var{a}x^{\\var{b}}+\\var{f}\\)   and   \\(v=\\var{c}cos(\\var{d}x)\\)

\n

then   \\(\\frac{du}{dx}=\\var{b}*\\var{a}x^{\\var{b}-1}\\)   and   \\(\\frac{dv}{dx}=-\\var{d}*\\var{c}sin(\\var{d}x)\\)

\n

Putting 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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\\(\\frac{df}{dx} = \\) [[0]]

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Differentiate the function:

\n

\\(f(x)=\\frac{\\var{a}x^{\\var{b}}+\\var{f}}{\\var{c}cos(\\var{d}x)}\\)

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Quotient rule

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question"}, {"name": "Turning points", "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": {"css": "", "js": ""}, "parts": [{"type": "gapfill", "marks": 0, "prompt": "

Input the smaller of the two \\(x\\) values. 

\n

\\(x=\\) [[0]]

\n

Input the larger of the two \\(x\\) values.

\n

\\(x=\\) [[1]]

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The function \\(f(x)=2x^3-\\simplify{3*({a}+{b})x^2+6*{a}*{b}x}+\\var{c}\\)  has two turning points.

", "ungrouped_variables": ["a", "b", "c"], "advice": "

\\(f(x)=2x^3-\\simplify{3*({a}+{b})x^2+6*{a}*{b}x}+\\var{c}\\)

\n

To 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\\)

\n

Divide across by 6 to get the quadratic equation

\n

\\(x^2-\\simplify{({a}+{b})x+{a}*{b}}=0\\)

\n

This 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}\\)

", "metadata": {"description": "

Turning points of a cubic function

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\\(f(x)=\\var{a1}e^{\\var{a2}x^{\\var{a3}}+\\var{a4}}\\)

\n

Recall the chain rule:   \\(\\frac{df}{dx}=\\frac{df}{du}.\\frac{du}{dx}\\)

\n

let \\(u=\\var{a2}x^{\\var{a3}}+\\var{a4}\\)    then   \\(f(x)=\\var{a1}e^{u}\\)

\n

\\(\\frac{df}{du}=\\var{a1}e^{u}\\)  and  \\(\\frac{du}{dx}=\\var{a3}*\\var{a2}x^{\\var{a3}-1}\\)

\n

\\(\\frac{df}{dx}=\\var{a1}e^{u}.\\simplify{{a2}*{a3}x^{{a3}-1}}\\)

\n

\\(\\frac{df}{dx}=\\simplify{{a1}*{a2}*{a3}x^{{a3}-1}}e^{\\var{a2}x^{\\var{a3}}+\\var{a4}}\\)

\n

", "rulesets": {}, "parts": [{"prompt": "

\\(\\frac{df}{dx}=\\) [[0]]

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Differentiate the function:

\n

\\(f(x)=\\var{a1}e^{\\var{a2}x^{\\var{a3}}+\\var{a4}}\\)

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Chain rule

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Rate of change problem involving velocity & acceleration

"}, "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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\\(h=\\var{a}t-4.9t^2\\)

\n

Recall 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\\)

\n

when \\(t=\\var{b}\\)

\n

\\(v=\\var{a}-2*4.9*\\var{b}\\)

\n

\\(v=\\simplify{{a}-9.8*{b}}m/s\\)

\n

\n

The 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\\)

\n

The 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}\\)

\n

\n

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Calculate the speed of the missile (m/s) \\(\\var{b}\\) seconds after launch. Give your answer correct to one decimal place.

\n

\\(v = \\) [[0]]m/s

\n

What is the maximum height achieved by this missile? Give your answer correct to one decimal place.

\n

\\(h = \\) [[1]]m

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\\(\\pi r^2h=\\var{v}\\)

\n

\\(h=\\frac{\\var{v}}{\\pi r^2}\\)

\n

The total surface area is to be a minimum.

\n

Lid + curved surface area + base

\n

\\(A=\\pi r^2+2\\pi rh+\\pi r^2\\)

\n

\\(A=2\\pi r^2+2\\pi r\\left(\\frac{\\var{v}}{\\pi r^2}\\right)\\)

\n

\\(A=2\\pi r^2+\\simplify{2*{v}}r^{-1}\\)

\n

\\(\\frac{dA}{dr}=4\\pi r-\\simplify{2{v}}r^{-2}=0\\)

\n

\\(4\\pi r=\\simplify{2*{v}}/{r^2}\\)

\n

\\(r^3=\\frac{\\var{v}}{2\\pi}\\)

\n

\\(r=\\simplify{({v}/(2*pi))^(1/3)}\\)

\n

From the second line we have the relation \\(h=\\frac{\\var{v}}{\\pi r^2}\\) to get

\n

\\(h=2*\\simplify{({v}/(2*pi))^(1/3)}\\)

\n

", "rulesets": {}, "parts": [{"prompt": "

Input the cyinder height, correct to two decimal places.

\n

\\(h = \\) [[0]]

\n

Input the required cylinder radius, correct to two decimal places.

\n

\\(r = \\) [[1]]

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A closed cylindrical tank is to be built having a volume of \\(\\var{v}\\) cc.

\n

Determine the required height, \\(h\\), and radius, \\(r\\), if the total surface area is to be a minimum.

\n

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Problem on a closed cylindrical tank having minimum surface area

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question"}, {"name": "Maximum/minimum", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Untitled.jpg", "/srv/numbas/media/question-resources/Untitled.jpg"], ["question-resources/Untitled_qCawkyB.jpg", "/srv/numbas/media/question-resources/Untitled_qCawkyB.jpg"]], "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": ["l", "w"], "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

The length of the box is \\(\\var{l}-2x\\), the width is \\(\\var{w}-2x\\) and the height is \\(x\\).

\n

The volume is then given by

\n

\\(V=(\\var{l}-2x).(\\var{w}-2x).x\\)

\n

\\(V=\\simplify{4x^3-2*({w}+{l})x^2+{l}*{w}x}\\)

\n

\\(\\frac{dV}{dx}=\\simplify{12x^2-4*({w}+{l})x+{l}*{w}}\\)

\n

This is a quadratic equation.

\n

\\(x=\\frac{\\simplify{4*({w}+{l})}\\pm\\sqrt(\\simplify{16*({w}+{l})^2-48*{w}*{l}})}{24}\\)

\n

\\(x=\\frac{\\simplify{{w}+{l}}\\pm\\sqrt(\\simplify{{w}^2-{w}*{l}+{l}^2})}{6}\\)

\n

\\(\\frac{d^2V}{dx^2}=\\simplify{24x-4*({w}+{l})}\\)

\n

when  \\(x=\\simplify{({w}+{l}-sqrt({w}^2-{w}*{l}+{l}^2))/6}\\)           \\(\\frac{d^2V}{dx^2}<0\\)      and therefore is the value that gives a maximum.

", "rulesets": {}, "parts": [{"prompt": "

Input the value for \\(x\\) correct to one decimal place.

\n

\\(x = \\) [[0]]

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A rectangular sheet of metal of length = \\(\\var{l}cm\\) and width  = \\(\\var{w}cm\\) has a square of side \\(x\\,cm\\) cut from each corner. The ends and sides will be folded upwards to form an open box.

\n

Determine the value of \\(x\\) that will maximise the volume of this box.

\n

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Maximising the volume of a rectangular box

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