3^x+log2(xy)=10
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Originally Posted by simsima_1 3^x+log2(xy)=10 for the first term, note that $\displaystyle \frac d{dx}a^x = a^x \ln a$ for $\displaystyle a > 0$ for the second part, use the change of base formula: $\displaystyle \log_ab = \frac {\ln b}{\ln a}$
Last edited by Jhevon; Feb 10th 2008 at 02:13 PM.
lol...ummm i think i'm more confused now ....
Originally Posted by simsima_1 lol...ummm i think i'm more confused now .... $\displaystyle 3^x + \log_2(xy) = 10$ $\displaystyle \Rightarrow 3^x + \frac {\ln (xy)}{\ln 2} = 10$ now continue by implicit differentiation
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