// Numbas version: exam_results_page_options {"name": "Maths Support: Questions on differentiation", "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "preventleave": false, "browse": true, "showfrontpage": false, "showresultspage": "never"}, "duration": 0.0, "metadata": {"notes": "", "description": "

Diagnostic testing on differentiation.

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$f(x)=\\var{a}x^{\\var{b}}$

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$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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Input all numbers as fractions or integers, not as decimals.

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$f(x)=\\var{a}x^{\\var{b}}+\\var{c}x^{1/\\var{d}}$

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$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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Input all numbers as fractions or integers, not as decimals.

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$\\displaystyle f(x)=\\frac{\\var{a}}{x^{1/\\var{c}}}+ \\frac{\\var{b}}{x^{-1/\\var{d}}}+\\var{ee}$

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$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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Input all numbers as fractions or integers, not as decimals.

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Differentiate the following functions $f(x)$ with respect to $x$.

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Input all numbers as fractions or integers, not as decimals.

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Differentiate $\\displaystyle ax^b, ax^b+cx^{1/d}, \\frac{a}{x^{1/c}}+\\frac{b}{x^{-1/d}}+c$ with respect to $x$.

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We have \$\\frac{df}{dx}=\\var{a*b}x^{\\var{b-1}}-\\var{c*d}x^{\\var{-d-1}}\$

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The gradient at $x=\\var{g}$ is given by the value of $\\displaystyle \\frac{df}{dx}$ at $x=\\var{g}$ and we therefore have:

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Gradient = $\\var{a*b}\\times(\\var{g})^{\\var{b-1}}-\\var{c*d}\\times (\\var{g})^{\\var{-d-1}}= \\var{dpformat(ans1,3)}$ to 3 decimal places.

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\$f(x) = \\simplify{ {a}*x^{b} + {c}/(x^{d}) + {ee}} \$

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Gradient at $x=\\var{g}\\;$ is [[0]]

\n

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Find the gradient of the curve $y= f(x)$ at the given point.

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Find the gradient of  $\\displaystyle ax^b+\\frac{c}{x^{d}}+f$ at $x=a$

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

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

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

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

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

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

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

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

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

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

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

\n

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

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

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Find the coordinates of the stationary points of the function and classify their types:

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

\n

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Find the stationary points of a cubic which has 2 turning points.

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c) Let $\\displaystyle \\simplify[std]{u={u}*e^({c}x)+{1-u}*ln({c}*x)}$

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So we have: $\\displaystyle \\frac{du}{dx}=\\simplify[std]{{u*c}*e^({c}x)+{1-u}/x}$ and $\\displaystyle f(u)=\\simplify[std]{{t[0]}*sin(u)+{t[1]}*cos(u)}\\Rightarrow \\frac{df}{du}= \\simplify[std]{{t[0]}*cos(u)-{t[1]}*sin(u)}$.

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By the Chain Rule \$\\begin{eqnarray*} \\frac{df}{dx}&=&\\frac{df}{du}\\times\\frac{du}{dx}\\\\&=&\\simplify[std]{{t[0]}*cos(u)-{t[1]}*sin(u)}\\times\\simplify[std]{{u*c}*e^({c}x)+{1-u}/x}\\\\&=&\\simplify[std]{{t[0]}*({u}*{c}e^({c}x)+{1-u}/x)*cos({u}*e^({c}x)+{1-u}*ln({c}*x))-{t[1]}*({u}*{c}e^({c}x)+{1-u}/x)*sin({u}*e^({c}x)+{1-u}*ln({c}*x))}\\end{eqnarray*}\$

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$\\displaystyle f(x)=\\simplify[all]{(x^{a}-{b})/(x^{-c}+{d})}$

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$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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Input all numbers as fractions or integers, not as decimals.

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$\\displaystyle f(x)=\\simplify[std]{(e^({a}x)-e^({b}x))/( e^({c}x)+e^({d}x))}$

\n

$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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Input all numbers as fractions or integers, not as decimals.

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$f(x)=\\simplify[std]{{t[0]}*sin({u}*e^({c}x)+{1-u}*ln({c}*x))+{t[1]}*cos({u}*e^({c}x)+{1-u}*ln({c}*x))}$.

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$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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Differentiate the following functions $f(x)$ with respect to $x$.

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Input all numbers as fractions or integers, not as decimals.

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Examples on differentiation using the quotient rule and chain rule.

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