If \(y = \log_e(\sec e^{x^2})\), then \(\displaystyle\frac{dy}{dx} =\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2xe^{x^2}\tan e^{x^2}\)
By the chain rule, working from the outside in (log, then sec, then \(e^{u}\), then \(x^2\)):
\(\displaystyle \frac{dy}{dx} = \frac{1}{\sec e^{x^2}} \cdot \sec e^{x^2}\tan e^{x^2} \cdot e^{x^2} \cdot 2x\)
\(= 2x\,e^{x^2}\tan e^{x^2}\)
Why the other options are incorrect
Option (1) \(x^2e^{x^2}\tan e^{x^2}\) — Incorrect. It has the factor \(x^2\) instead of the derivative of \(x^2\), which is \(2x\).
Option (3) \(e^{x^2}\tan e^{x^2}\) — Incorrect. It leaves out the last chain-rule factor \(\frac{d}{dx}(x^2) = 2x\).
Option (4) \(xe^{x^2}\tan e^{x^2}\) — Incorrect. The last factor is \(2x\), not x.