// Numbas version: finer_feedback_settings {"name": "Musa's copy of 3 Chain rule - log of binomial", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Musa's copy of 3 Chain rule - log of binomial", "tags": [], "metadata": {"description": "
Differentiate $\\displaystyle \\ln((ax+b)^{m})$
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the chain rule.
", "advice": "$\\simplify[std]{f(x) = ln({a}x+{b}x^{m})}$
\nThe chain rule applies to $f(x)=g(h(x))$ where
\n\\[ g(h) = \\ln(h) {\\rm~and~} \\simplify[std]{h(x) = {a}x+{b}x^{m}}.\\]
\nThen we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{dh}{dx} \\cdot \\frac{dg}{dh}\\]
Calculate the derivative of $h(x)$ and $g(h)$: \\[\\frac{dh}{dx} = \\simplify[std]{{a}+{b*m}x^{m-1}}\\]
\n\\[\\frac{dg}{dh} = \\frac{1}{h}\\]
\nHence on substituting $h = h(x) = \\simplify[std]{{a}x+{b}x^{m}}$ we finally have
\n\\[\\frac{df}{dx} = \\simplify[std]{({a}x+{b*m}x^{m-1})/({a}x+{b}x^{m})} \\]
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\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
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