Differentiate the following function. $f(x) = -8 + 3x + 2e^x$ $f'(x) = \boxed{}$
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The derivative of a constant is 0. So, $\frac{d}{dx}(-8) = 0$. The derivative of $cx$ where c is a constant is c. So, $\frac{d}{dx}(3x) = 3$. The derivative of $ce^x$ where c is a constant is $ce^x$. So, $\frac{d}{dx}(2e^x) = 2e^x$. Show more…
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