Differentiate the following function $f(x)$ using the chain rule.
", "name": "Luis's copy of Chain rule - trig and quadratic,", "metadata": {"notes": "1/08/2012:
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", "licence": "Creative Commons Attribution 4.0 International", "description": "Differentiate $\\displaystyle \\cos(e^{ax}+bx^2+c)$
"}, "ungrouped_variables": ["a", "s2", "c", "b", "s1"], "parts": [{"prompt": "\\[\\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}\\]
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
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", "type": "gapfill", "marks": 0, "showCorrectAnswer": true, "gaps": [{"checkvariablenames": false, "showCorrectAnswer": true, "answer": "-({a}e^({a}x)+{2*b}x)*sin(e^({a}x) +{b}x^2+{c})", "vsetrange": [0, 1], "answersimplification": "std", "expectedvariablenames": [], "vsetrangepoints": 5, "type": "jme", "marks": 3, "checkingtype": "absdiff", "scripts": {}, "checkingaccuracy": 0.001, "showpreview": true}], "stepsPenalty": 0, "scripts": {}, "steps": [{"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 $x$.
$\\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}$
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 $u=\\simplify[std]{e^({a}x) +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{cos(u)}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a}e^({a}x) +{2*b}x}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{-sin(u)} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({a}e^({a}x) +{2*b}x) * (-sin(u))}\\\\\n \n &=& \\simplify[std]{-({a}e^({a}x) +{2*b}x)*sin(e^({a}x) +{b}x^2+{c})}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{e^({a}x) +{b}x^2+{c}}$.