Here is a problem that I am unsure how to proceed.. Differentiate 3^x
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$\displaystyle (a^x)'=a^x\cdot\ln a$
Hi Originally Posted by dankelly07 Here is a problem that I am unsure how to proceed.. Differentiate 3^x This is simple, because $\displaystyle 3^x = e^{x*ln(3)}$ $\displaystyle [e^{x*ln(3)}] ' = ln(3) * [e^{x*ln(3)}] = ln(3) *3^x$
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