Derivative of a sum of log, exponential, and power terms

Apply the sum/difference and constant multiple rules to differentiate term by term: $\frac{d}{dx}[8\ln x] = 8\cdot\frac{1}{x} = \frac{8}{x}$, $\frac{d}{dx}[-3e^x] = -3e^x$, and $\frac{d}{dx}[x^4] = 4x^3$ by the power rule. Combining these gives $f'(x) = \frac{8}{x} - 3e^x + 4x^3$.

Now substitute $x = 1$: $f'(1) = \frac{8}{1} - 3e^{1} + 4(1)^3 = 8 - 3e + 4 = 12 - 3e$. So the final answer is $f'(1) = 12 - 3e$.