Step 1: Use logarithmic differentiation.
Take the natural log of \(y=e^{4x}92^x\). Products turn into sums, which are easier to handle.
\[ \ln y=4x+x\ln 92 \]
Step 2: Differentiate both sides.
On the left, \(\frac{d}{dx}\ln y=\frac{1}{y}\frac{dy}{dx}\). On the right, \(\ln 92\) is a constant, so the right side gives \(4+\ln 92\).
\[ \frac{1}{y}\frac{dy}{dx}=4+\ln 92 \]
Step 3: Multiply back by y.
\[ \frac{dy}{dx}=y\,(4+\ln 92)=e^{4x}(92)^x\,(4+\ln 92) \]
Step 4: Match with the options.
The question says \(\log_e x=\log x\), so \(\ln 92=\log 92\) here. The result equals option 3. The other options have a factor like \(x\) or \(92^{x-1}\), which only appear when the exponent is wrongly treated as a fixed power.
Final Answer:
So \(\frac{dy}{dx}=e^{4x}(92)^x[4+\log 92]\).
\[ \boxed{e^{4x}(92)^x\,[4+\log 92]} \]