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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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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.

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\\(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}\\)

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Turning points of a cubic function

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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

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Rate of change problem involving velocity & acceleration

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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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