\\[\\simplify[std]{f(x) = ln(({a}x+{b})^{m})}\\]
\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": [{"showCorrectAnswer": true, "marks": 0, "scripts": {}, "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.
", "metadata": {"description": "Differentiate $\\displaystyle \\ln((ax+b)^{m})$
", "notes": "\n\t\t\n\t\t
1/08/2012:
\n\t\tAdded tags.
\n\t\tAdded description.
\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\tChecking range chosen so that the denominator of the result is never 0.
\n\t\t\n\t\t", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "functions": {}, "variable_groups": [], "question_groups": [{"questions": [], "pickQuestions": 0, "name": "", "pickingStrategy": "all-ordered"}], "showQuestionGroupNames": false, "advice": "\n\t \n\t \n\t
$\\simplify[std]{f(x) = ln(({a}x+{b})^{m})}$
First note that we can simplify this by using the rule that $\\simplify[std]{ln(a^r)=r*ln(a)}$.
Hence $\\simplify[std]{f(x) = ln(({a}x+{b})^{m})={m}ln({a}x+{b})}$
So we need to differentiate $\\simplify[std]{ln({a}x+{b})}$
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 $u=\\simplify[std]{{a}x +{b}}$ and we have $f(u)=\\simplify[std]{{m}*ln(u)}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a}}\\\\\n\t \n\t \\frac{df(u)}{du} &=& \\simplify[std]{{m}/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}/u)}\\\\\n\t \n\t &=& \\simplify[std]{{a*m}/({a}x+{b})}\n\t \n\t \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x+{b}}$.