how would i differentiate f(x) = 3e^x - 0.5 ln x - 2
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$\displaystyle f^{\prime}(x) = 3e^x - 0.5 \frac {1}{x-2}$
Originally Posted by sharp357 how would i differentiate f(x) = 3e^x - 0.5 ln x - 2 You should use the properties: $\displaystyle h(x) = e^x~\implies~h'(x) = e^x$ and $\displaystyle g(x)=\ln(x+a)~\implies~g'(x)=\dfrac1{x+a}, a\in \mathbb{R}$
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