![](\"resources/question-resources/Car_image.jpg\")
Its speed at $A$ is $\\var{u} \\mathrm{~m/s}$ and it takes $\\var{t}$ seconds to move from $A$ to $B$.\\[\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}\\]
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
", "steps": [{"type": "information", "prompt": "\n \n \nThe chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
Differentiate the following function $f(x)$ using the chain rule.
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\nAdded tags.
\nAdded description.
\nChecked calculation. OK.
\nAdded information about Show steps. Altered to 0 marks lost rather than 1.
\nGot rid of a redundant ruleset.
\nImproved display in prompt.
\n", "licence": "Creative Commons Attribution 4.0 International", "description": "
Differentiate $\\displaystyle (ax^m+b)^{n}$.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n \n \n$\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
For this example, we let $u=\\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\\simplify[std]{u^{n}}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{m*a}x ^ {m -1}}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{{n}u^{n-1}} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{{m*a}x ^ {m-1} * ({n}*u^{n-1})}\\\\\n \n &=&\\simplify[std]{{m*a*n}x^{m-1}u^{n-1}}\\\\\n \n &=& \\simplify[std]{{m*a*n}x^{m-1}({a}*x^{m}+{b})^{n-1}}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.
\\[\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}\\]
\n\t\t\t$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\n\t\t\tClick on Show steps for more information. You will not lose any marks by doing so.
\n\t\t\t", "steps": [{"type": "information", "prompt": "\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\tThe chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
Differentiate the following function $f(x)$ using the chain rule.
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\n\t\tAdded tags.
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\n\t\tChecked calculation. OK.
\n\t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n\t\tGot rid of a redundant ruleset.
\n\t\tImproved display in prompt.
\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Differentiate $\\displaystyle e^{ax^{m} +bx^2+c}$
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t \n\t \n\t$\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
For this example, we let $u=\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{e^u}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a*m}x^{m-1} +{2*b}x}\\\\\n\t \n\t \\frac{df(u)}{du} &=& \\simplify[std]{e^u} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n\t \n\t \n\t \n\t\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x) * (e^u)}\\\\\n\t \n\t &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x)*e^({a}x^{m} +{b}x^2+{c})}\n\t \n\t \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.
Differentiate $\\displaystyle \\ln((ax+b)^{m})$
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the chain rule.
", "advice": "$\\simplify[std]{f(x) = ln({a}x+{b}x^{m})}$
\nThe chain rule applies to $f(x)=g(h(x))$ where
\n\\[ g(h) = \\ln(h) {\\rm~and~} \\simplify[std]{h(x) = {a}x+{b}x^{m}}.\\]
\nThen we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{dh}{dx} \\cdot \\frac{dg}{dh}\\]
Calculate the derivative of $h(x)$ and $g(h)$: \\[\\frac{dh}{dx} = \\simplify[std]{{a}+{b*m}x^{m-1}}\\]
\n\\[\\frac{dg}{dh} = \\frac{1}{h}\\]
\nHence on substituting $h = h(x) = \\simplify[std]{{a}x+{b}x^{m}}$ we finally have
\n\\[\\frac{df}{dx} = \\simplify[std]{({a}x+{b*m}x^{m-1})/({a}x+{b}x^{m})} \\]
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\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following expressions with respect to $x$ using the product rule.
\nSimplify your answers as much as possible.
", "advice": "These questions use the product rule.
\nThe product rule is defined as
\n$\\frac{dy}{dx}=v\\frac{du}{dx}+u\\frac{dv}{dx}$
\nwhen $y=uv$
\nWorked example using Part a:
\n$y=\\simplify{(x+{c[0]})(x+{c1})}$
\nThis expression is the result of two products; $(\\simplify{x+{c[0]}})$ and $(\\simplify{x+{c1}})$.
\nWe can therefore say:
\n$u=(\\simplify{x+{c[0]}})$
\nand
\n$v=(\\simplify{x+{c1}})$,
\nHence meaning that $y=uv$.
\n\nWe already have what $u$ and $v$ equal, so all we have to do is find what $\\frac{du}{dx}$ and $\\frac{dv}{dx}$ are, and then substitute everything into the rule.
\nDifferentiating with respect to $x$, we get:
\n$\\frac{du}{dx}=1$
\nand
\n$\\frac{dv}{dx}=1$.
\nAs there are no powers or coefficients of $x$ that are $>1$, this is a very simple version of the product rule, but knowing how to work out this equation formally will make more difficult looking problems just as simple.
\nSubstituting in all the results we've found, we get:
\n$\\frac{dy}{dx}=1(\\simplify{x+{c1}})+1(\\simplify{x+{c[0]}})$
\nWe then simplify, collecting all the terms, to get our final answer of:
\n$\\frac{dy}{dx}=\\simplify{2x+{c[0]}+{c1}}$
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "\nA car is driving in a straight line from $A$ to $B$ with constant acceleration $\\var{a} \\mathrm{~m/s^{2}}$.
\nIts speed at $A$ is $\\var{u} \\mathrm{~m/s}$ and it takes $\\var{t}$ seconds to move from $A$ to $B$.
You start by writing down the values you know and the values you need to find.
\n$a = \\var{a},$ $u=\\var{u},$ $t=\\var{t},$ $v=$ $?,$ $s=$ $?$
\na) At $B$ the car will have reached its final velocity, $v \\mathrm{ms^{-1}}$. You need $v$ and you know $u, a$ and $t$ so you can use the equation $v= u +at$. \\begin{align} v &= u + at, \\\\
&= \\var{u} + (\\var{a} \\times \\var{t}), \\\\
&= \\var{u +a*t} \\mathrm{ms^{-1}}. \\end{align}
The speed of the car at $B$ is $\\var{u + a*t} \\mathrm{ms^{-1}}.$
\nb) You need the distance, $s \\mathrm{m}$. You calculated $v$ in the previous part, so you can use the equation $s=\\left(\\frac{u+v}{2}\\right)t.$
\n\\begin{align} s &= \\left(\\frac{u+v}{2}\\right)t, \\\\
&= \\left(\\frac{\\var{u}+\\var{v}}{2}\\right) \\times \\var{t}, \\\\
&= \\var{((u+v)/2)*t} \\mathrm{m}. \\end{align}
The distance from $A$ to $B$ is $\\var{((u+v)/2)*t}$ metres.
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