Any help with solving this... I can't figure it out. $\displaystyle 3^xlog_8(x)$ That is, I need to find the derivative. I think this is the first step: $\displaystyle ln(3)/ln(8) $
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$\displaystyle 3^x \log_8 x = \frac{3^x \ln x}{\ln 8}$
Do i continue with using the quetient rule. Is f'(x)= $\displaystyle 3^xln(x)*3$?
Originally Posted by cjmac87 Do i continue with using the quetient rule. Is f'(x)= $\displaystyle 3^xln(x)*3$? $\displaystyle \frac{3^x \ln{x}}{\ln{8}} = \frac{1}{\ln{8}} \cdot 3^x \ln{x} $ use the product rule ... $\displaystyle \frac{1}{\ln{8}}\left(3^x \cdot \frac{1}{x} + 3^x \cdot \ln{3} \cdot \ln{x}\right)$ now clean up the algebra.
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