If \( y=e^{4x}(92)^x \); then \( \displaystyle\frac{dy}{dx} \) is equal to
(\( \log_e x=\log x \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( e^{4x}(92)^x[4+\log 92] \)
By the product rule, with \( \frac{d}{dx}e^{4x}=4e^{4x} \) and \( \frac{d}{dx}a^x=a^x\log a \):
\( \displaystyle \frac{dy}{dx}=4e^{4x}(92)^x+e^{4x}(92)^x\log 92 \)
\( \displaystyle \frac{dy}{dx}=e^{4x}(92)^x[4+\log 92] \)
Shortcut: Take logarithms: \( \log y=4x+x\log 92 \), so \( \frac{1}{y}\frac{dy}{dx}=4+\log 92 \) and \( \frac{dy}{dx}=y(4+\log 92) \).
Why the other options are incorrect
Option (1) \( 4x\,e^{4x}.92^{x-1} \) — Incorrect. This treats \( 92^x \) like a power \( x^n \); the derivative of \( a^x \) is \( a^x\log a \), not \( xa^{x-1} \).
Option (2) \( e^{4x}92^{x-1}[368+x] \) — Incorrect. This also uses the power rule on \( 92^x \): \( 4\cdot 92\cdot 92^{x-1}+x\cdot 92^{x-1}=92^{x-1}(368+x) \), which is wrong for an exponential.
Option (4) \( e^{4x}92^x\log 92 \) — Incorrect. Only the derivative of \( 92^x \) has been taken; the term \( 4e^{4x}(92)^x \) from differentiating \( e^{4x} \) is missing.