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"], "functions": {}, "question_groups": [{"name": "", "pickQuestions": 0, "pickingStrategy": "all-ordered", "questions": []}], "variables": {"c": {"group": "Ungrouped variables", "definition": "s2*random(1..9)", "templateType": "anything", "description": "", "name": "c"}, "b": {"group": "Ungrouped variables", "definition": "s1*random(1..9)", "templateType": "anything", "description": "", "name": "b"}, "a": {"group": "Ungrouped variables", "definition": "random(2..9)", "templateType": "anything", "description": "", "name": "a"}, "m": {"group": "Ungrouped variables", "definition": "random(3..4)", "templateType": "anything", "description": "", "name": "m"}, "s1": {"group": "Ungrouped variables", "definition": "random(1,-1)", "templateType": "anything", "description": "", "name": "s1"}, "s2": {"group": "Ungrouped variables", "definition": "random(1,-1)", "templateType": "anything", "description": "", "name": "s2"}}, "name": "Luis's copy of Chain rule - exponential of polynomial,", "ungrouped_variables": ["a", "c", "b", "s2", "s1", "m"], "variable_groups": [], "type": "question", "preamble": {"css": "", "js": ""}, "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"description": "Differentiate $\\displaystyle e^{ax^{m} +bx^2+c}$
", "notes": "\n\t\t1/08/2012:
\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\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 0, "prompt": "\n\t\t\t\\[\\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 $x$.
$\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$
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 $\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.