\\[\\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.
", "tags": ["Calculus", "MAS1601", "SFY0004", "Steps", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"], "surdf": [{"result": "(sqrt(b)*a)/b", "pattern": "a/sqrt(b)"}]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "1/08/2012:
\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 $\\simplify[std]{{a}x^{m}+{b}}$.