// Numbas version: finer_feedback_settings {"name": "Luis's copy of Chain rule - binomial,", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"ungrouped_variables": ["a", "s1", "b", "m", "n"], "name": "Luis's copy of Chain rule - binomial,", "variablesTest": {"condition": "", "maxRuns": 100}, "preamble": {"js": "", "css": ""}, "parts": [{"prompt": "
\\[\\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.
", "gaps": [{"checkvariablenames": false, "showpreview": true, "vsetrangepoints": 5, "answersimplification": "std", "type": "jme", "expectedvariablenames": [], "scripts": {}, "marks": 3, "checkingaccuracy": 0.001, "checkingtype": "absdiff", "answer": "{a*m*n}x ^ {m-1} * ({a} * x^{m}+{b})^{n-1}", "showCorrectAnswer": true, "vsetrange": [0, 1]}], "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 $x$.
$\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$
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 $\\simplify[std]{{a}x^{m}+{b}}$.
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", "description": "
Differentiate $\\displaystyle (ax^m+b)^{n}$.
"}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}