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\\(\\frac{df}{dx}=\\) [[0]]

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Differentiate the function:

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\\(f(x)=\\var{a1}ln(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

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Chain rule

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\\(f(x)=\\var{a1}sin(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

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Recall the chain rule:   \\(\\frac{df}{dx}=\\frac{df}{du}.\\frac{du}{dx}\\)

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let \\(u=\\var{a2}x^{\\var{a3}}+\\var{a4}\\)    then   \\(f(x)=\\var{a1}sin(u)\\)

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\\(\\frac{df}{du}=\\var{a1}cos(u)\\)  and  \\(\\frac{du}{dx}=\\var{a3}*\\var{a2}x^{\\var{a3}-1}\\)

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\\(\\frac{df}{dx}=\\var{a1}cos(u).\\simplify{{a2}*{a3}x^{{a3}-1}}\\)

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\\(\\frac{df}{dx}=\\simplify{{a1}*{a2}*{a3}x^{{a3}-1}}cos(\\var{a2}x^{\\var{a3}}+\\var{a4})\\)

", "type": "question", "contributors": [{"name": "Milena Venkova", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2169/"}]}]}], "contributors": [{"name": "Milena Venkova", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2169/"}]}