Differentiate the following function $f(x)$ using the chain rule.
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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}{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) = ln(({a}x+{b})^{m})}\\]
\n\t\t\t$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
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\n\t\t\t", "stepsPenalty": 0}], "tags": ["Calculus", "MAS1601", "Steps", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function", "logarithm laws", "logarithms"], "advice": "\n\t \n\t \n\t$\\simplify[std]{f(x) = ln(({a}x+{b})^{m})}$
First note that we can simplify this by using the rule that $\\simplify[std]{ln(a^r)=r*ln(a)}$.
Hence $\\simplify[std]{f(x) = ln(({a}x+{b})^{m})={m}ln({a}x+{b})}$
So we need to differentiate $\\simplify[std]{ln({a}x+{b})}$
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 $u=\\simplify[std]{{a}x +{b}}$ and we have $f(u)=\\simplify[std]{{m}*ln(u)}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a}}\\\\\n\t \n\t \\frac{df(u)}{du} &=& \\simplify[std]{{m}/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}/u)}\\\\\n\t \n\t &=& \\simplify[std]{{a*m}/({a}x+{b})}\n\t \n\t \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x+{b}}$.