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Remember the rule:

\n

If $y = ax^n$ then $\\frac{dy}{dx} = anx^{n-1}$

\n

\n

Differentiate each of the following:

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i)

\n

$y = x^{\\var{pow[0]}}$

\n

$\\frac{dy}{dx} = \\var{ans11}x^{\\var{pow[0]}-1} = \\var{ans11}x^{\\var{ans12}}$

\n

ii)

\n

$y = \\var{num[0]}x^{\\var{pow[1]}}$

\n

$\\frac{dy}{dx} = \\var{ans21}x^{(\\var{pow[1]}-1)} = \\var{ans21}x^{\\var{ans22}}$

\n

iii)

\n

$y = \\var{num[1]}x^{\\var{pow[2]}}$

\n

$\\frac{dy}{dx} = \\var{ans31}x^{(\\var{pow[2]}-1)} = \\var{ans31}x^{\\var{ans32}}$

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$y = x^{\\var{pow[0]}}$

\n

$\\frac{dy}{dx} =$ [[0]]

\n

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$y = \\var{num[0]}x^{\\var{pow[1]}}$

\n

$\\frac{dy}{dx} =$ [[0]]

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$y = \\var{num[1]}x^{\\var{pow[2]}}$

\n

$\\frac{dy}{dx} =$ [[0]]

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Remember the rule:

\n

If $y = ax^n$ then $\\frac{dy}{dx} = anx^{n-1}$

\n

\n

Differentiate each of the following:

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i)

\n

$y = x^{\\frac{1}{\\var{p1}}}$

\n

$\\frac{dy}{dx} = \\var{ans11}x^{(\\frac{1}{\\var{p1}}-1)} = \\var{ans11}x^{\\var{ans12}}$

\n

ii)

\n

$y = x^{\\frac{\\var{p2[0]}}{\\var{p2[1]}}}$

\n

$\\frac{dy}{dx} = \\frac{\\var{p2[0]}}{\\var{p2[1]}}x^{(\\frac{\\var{p2[0]}}{\\var{p2[1]}}-1)} = \\var{ans21}x^{\\var{ans22}}$

\n

iii)

\n

$y = \\var{num[0]}x^{\\var{p3}}$

\n

$\\frac{dy}{dx} = \\var{ans31}x^{(\\var{p3}-1)} = \\var{ans31}x^{\\var{ans32}}$

\n

iv)

\n

$y = \\var{num[1]}\\sqrt(x) = \\var{num[1]}x^{\\frac{1}{2}}$

\n

$\\frac{dy}{dx} = \\frac{\\var{num[1]}}{2}x^{(\\frac{1}{2}-1)} = \\var{ans41}x^{\\var{ans42}}$

\n

v)

\n

$y = \\frac{\\var{n51}}{x^{\\var{n52}}} = \\var{n51}x^{-\\var{n52}}$

\n

$\\frac{dy}{dx} = \\var{ans51}x^{(-\\var{n52}-1)} = \\var{ans51}x^{\\var{ans52}}$

\n

vi)

\n

$y = \\sqrt(x^\\var{n6}) = x^{\\frac{\\var{n6}}{2}}$

\n

$\\frac{dy}{dx} = \\frac{\\var{n6}}{2}x^{(\\frac{\\var{n6}}{2}-1)} = \\var{ans61}x^{\\var{ans62}}$

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$y = \\var{num[0]}x^{\\var{p3}}$

\n

$\\frac{dy}{dx} =$ [[0]]

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$y = \\var{num[1]}\\sqrt x$

\n

$\\frac{dy}{dx} =$ [[0]]

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Hint: What is $\\sqrt x$ as a power of $x$

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$y = \\frac{\\var{n51}}{x^{\\var{n52}}}$

\n

$\\frac{dy}{dx} =$ [[0]]

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Hint: What is $\\frac{1}{x^2}$ as a power of $x$

{eqnline(a,b,x2,y2)}

\n

The above graph shows a graph of a quadratic equation, it is your task to find this equation.

\n

You are given the two points of the curve with the x axis, $(\\var{b},0)$ and $(\\var{a},0)$, and the $y$-intercept at $(0,\\var{c})$ as indicated on the diagram.

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Write the equation of the graph in the diagram.

\n

$y=\\;$[[0]]

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Find the coordinates of the turning point of this quadratic

\n

$x=$[[0]]

\n

$y=$[[1]]

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Students enter equation and turning point

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We know that the graph crosses the $x$-axis at both $(\\var{a},0)$ and $(\\var{b},0)$. Since this is a quadratic, we know our equations has two roots, and by the previous observation, they are at $\\var{a}$ and $\\var{b}$. Hence we can write our equation as $\\simplify{y=(x-{a})(x-{b})}$ which simplifies to $\\simplify{y=x^2-({a}+{b})x+({a}*{b})}$.

\n

\n

To find the coefficients of the turning point of the quadratic, we know the x-coordinate of the turning point will correspond to the solution to $dy/dx=0$. So we get $\\simplify{2x-({a}+{b})}=0$ hence $\\simplify{x=({a}+{b})/2}$. We substitute this value of x back into the equation of the quadratic to find the corresponding y-coordinate.

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On differentiating we get $\\displaystyle \\frac{df}{dx}=\\simplify[std]{{3*a}x^2+{2*b}x+{c}}$.

\n

To find the stationary points we have to solve $\\displaystyle \\frac{df}{dx}=0$ for $x$.

\n

So we have to solve $\\simplify[std]{{3*a}x^2+{2*b}x+{c}=0}$.

\n

Note that the quadratic factorises and the equation becomes $\\simplify[std]{({3a}x-{r1})(x-{r2})=0}$.

\n

Hence we have two stationary points: $x=\\simplify[std]{{r1}/{3a}}$ and $x=\\var{r2}$.

\n

To find out the types of these stationary points we look at the sign of $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6a}*x+{2*b}}$ at  the stationary points.

\n

If  $\\displaystyle \\frac{d^2f}{dx^2} \\lt 0$ at a stationary point then it is a MAXIMUM.

\n

If  $\\displaystyle \\frac{d^2f}{dx^2} \\gt 0$ at a stationary point then it is a MINIMUM.

\n

If  $\\displaystyle \\frac{d^2f}{dx^2} = 0$ at a stationary point then we have to do more work!

\n

At $x=\\var{r2}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6*a*r2+2*b}}${lg1}$0$ hence is a {type1}.

\n

At $\\displaystyle x=\\simplify[std]{{r1}/{3a}}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{2*r1+2*b}}${lg2}$0$ hence is a {type2}.

\n

", "parts": [{"type": "gapfill", "prompt": "

$f(x)=\\simplify[all,!collectNumbers,!noleadingminus]{{a}x^3+{b}x^2+{c}x+{d}}$

\n

$f'(x)=$ [[2]]

\n

$f''(x)=$ [[3]]

\n

\n

Find when $f'(x)=0$, hence find:

\n

$x$-coordinate of the stationary point giving a minimum $=$ [[0]]

\n

$x$-coordinate of the stationary point giving a maximum $=$ [[1]]

Finding the stationary points of a cubic with two turning points

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Find the coordinates of the stationary points of the function.

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Input the smaller of the two \$$x\$$ values.

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\$$x=\$$ [[0]]

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Input the larger of the two \$$x\$$ values.

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\$$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}\$$

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To locate a turning point, differentite the function, set equal to zero and solve.

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\$$f'(x)=6x^2-\\simplify{6*({a}+{b})x+6*{a}*{b}}=0\$$

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Divide across by 6 to get the quadratic equation

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

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This has factors

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\$$(x-\\var{a})(x-\\var{b})=0\$$

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\$$x-\\var{a}=0\$$     or     \$$x-\\var{b}=0\$$

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\$$x=\\var{a}\$$     or     \$$x=\\var{b}\$$

Turning points of a cubic function

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Parts A and B

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Here, the question takes you throught the stages needed to find the solution. The reason we differentiate is that the derivative of a function tells us its gradient at a given point, and we want to find where the function has gradient zero because when the gradient is zero we either have a maximum or a minimum point.

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Part C

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The first part of this question is similar to parts A and B. The tricky bit is the second part! You need to work out the value of $t$ that produces the maximum piont but that is not the final answer - you need to use that value of $t$ to find the maximum height, which you do by substituting $t$ into the original function to find $y$.

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Is the stationary point a maximum?

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Find the gradient of the curve $y$ at the point $x=\\var{d}$, giving your answer to $2$ decimal places if necessary.

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\$y = \\simplify{ {a}*x^2 + {b}x + {c}} \$

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Firstly, differentiate.

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$\\displaystyle \\frac{dy}{dx}=$ [[1]]

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Gradient at $x=\\var{d}\\;$ is [[0]]

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Find the coordinates of the turning point of the function below and state whether it is a maximum or a minimum point. Give your answers to $2$ decimal places where necessary.

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$y=\\simplify {{f}x^2+{g}x+{h}}$

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Firstly, find the first and second derivatives $y$.

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$\\displaystyle \\frac{dy}{dx}=$ [[2]]

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$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]

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Secondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.

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$x$-coordinate of the turning point $=$ [[0]]

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$y$-coordinate of the turning point $=$ [[1]]

\n

The turning point is a [[4]]

\n

\n

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maximum

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minimum

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An unpowered missile is launched vertically from the ground.

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At a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula

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\$y=\\var{z}t-\\var{w}t^2. \$

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Calculate the maximum height reached by the missile.

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Firstly, differentiate.

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$\\displaystyle \\frac{dy}{dt}=$ [[0]]

\n

Now use this result and your knowledge of differentiation to find the maximum height of the missile, rounding your answer to $2$ decimal places.

\n

$y=$ [[1]]

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You have 50 minutes to complete this exam. You may pause it at any time. Please ensure you enter this exam on one device only. The pass rate is 90% and you have unlimited attempts. However once you have passed the exam do not re enter it or you will wipe your score.

\n

Exam is due to be completed by Sunday 22nd April at 23:00 and is worth 2%.

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Assesses basic differentiation and finding turning points of curves.

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