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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

\$$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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\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

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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}\$$

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

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\$$\\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

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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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\$$\\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

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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

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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

\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.

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

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