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Differentiate $\\displaystyle \\ln((ax+b)^{m})$

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Differentiate the following function $f(x)$ using the chain rule.

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$\\simplify[std]{f(x) = ln({a}x+{b}x^{m})}$

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The chain rule applies to $f(x)=g(h(x))$ where

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\\[ g(h) = \\ln(h) {\\rm~and~} \\simplify[std]{h(x) = {a}x+{b}x^{m}}.\\]

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Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{dh}{dx} \\cdot \\frac{dg}{dh}\\]

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Calculate the derivative of $h(x)$ and $g(h)$: \\[\\frac{dh}{dx} = \\simplify[std]{{a}+{b*m}x^{m-1}}\\]

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\\[\\frac{dg}{dh} = \\frac{1}{h}\\]

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Hence on substituting $h = h(x) = \\simplify[std]{{a}x+{b}x^{m}}$ we finally have

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\\[\\frac{df}{dx} = \\simplify[std]{({a}x+{b*m}x^{m-1})/({a}x+{b}x^{m})}    \\]

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\\[\\simplify[std]{f(x) = ln({a}x+{b}x^{m})}\\]

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$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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Click on Show steps for more information. You will not lose any marks by doing so.

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