Derivar 7^x+1/ln (x+1).... Pergunta de ideia deMhary11

December 2019 1 85 Report

Derivar 7^x+1/ln (x+1)

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Comentários (5) Será aplicada algumas propriedades de derivação de funções exponenciais:

$f'(x)=7^{\frac{x+1}{\ln(x+1)}} \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{x+1}{\ln(x+1)}.\ln(7)]' \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[(\frac{x+1}{\ln(x+1)})'.\ln(7)+(\frac{x+1}{\ln(x+1)}).\ln(7)'] \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[(\frac{x+1}{\ln(x+1)})'.\ln(7)+(\frac{x+1}{\ln(x+1)}).0] \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{(x+1)'\ln(x+1)-(x+1)\ln(x+1)'}{[\ln(x+1)]^2}.\ln(7)]$

$f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{1.\ln(x+1)-(x+1).\frac{1}{(x+1)}.(x+1)'}{[\ln(x+1)]^2}.\ln(7)] \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{\ln(x+1)-(x+1).\frac{1}{(x+1)}.1}{[\ln(x+1)]^2}.\ln(7)] \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{\ln(x+1)-(x+1).\frac{1}{(x+1)}}{[\ln(x+1)]^2}.\ln(7)] \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{\ln(x+1)-\frac{x+1}{x+1}}{[\ln(x+1)]^2}.\ln(7)]$

$f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{\ln(x+1)-1}{\ln^2(x+1)}.\ln(7)] \\ \\ f'(x)=7^{\frac{x+1}{\ln(x+1)}}.[\frac{\ln(7)\ln(x+1)-1}{\ln^2(x+1)}] \\ \\ \left[\begin{array}{ccc}f'(x)=\frac{\ln(7)7^{\frac{x+1}{\ln(x+1)}}\ln(x+1)-1}{\ln^2(x+1)}\end{array}\right]$

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