1. Statement of the problem. We are asked to find the derivative of the function $f(x)=3x+\frac{4}{(x^2+3x+5)^2}$. 2. Formula and important rules used. We use the sum rule, the power rule, and the chain rule. If $f(x)=u(x)^n$ then $$\frac{d}{dx}[u(x)^n]=n u(x)^{n-1} u'(x)$$ Also $\frac{d}{dx}[u+v]=u'+v'$. 3. Rewrite the function to apply the power rule. We rewrite $f(x)=3x+4(x^2+3x+5)^{-2}$. 4. Differentiate term by term. The derivative of $3x$ is $3$. For the second term apply the chain rule with $u(x)=x^2+3x+5$ and $n=-2$. Compute the derivative step by step: $$f'(x)=3+4\cdot(-2)(x^2+3x+5)^{-3}\cdot(2x+3)$$ 5. Simplify the expression. Multiply the constants to get: $$f'(x)=3-8(2x+3)(x^2+3x+5)^{-3}$$ Write as a single fraction: $$f'(x)=3-\frac{8(2x+3)}{(x^2+3x+5)^3}$$ 6. Final answer. Therefore the derivative is $f'(x)=3-\frac{8(2x+3)}{(x^2+3x+5)^3}$. 7. Domain note. Since the quadratic $x^2+3x+5$ has discriminant $\Delta=3^2-4\cdot1\cdot5=-11<0$, it is never zero, so the function and its derivative are defined for all real numbers.