Question

If \(y= log(sec\ e^{x^2})\), then \(\frac {dy}{dx}\) =?

• A. \(x^2e^{x^2} tan\ e^2\)
• B. \(e^{x^2}.tan\ e^{x^2}\)
• C. \(2x.e^{x^2}.tan\ e^{x^2}\)
• D. \(x.e^{x^2} tan\ e^{x^2}\)

Answer 1

The Correct Option is C

Let \(y = log(sec\ e^{x^2})\)
\(\frac {dy}{dx}= \frac {d}{dx} [log(sec\ e^{x^2})]\)

\(\frac {dy}{dx}=\) \(\frac {1}{sec(e^{x^2})}.\frac {d}{dx} [sec\ e^{x^2}]\)

\(\frac {dy}{dx}=\) \(\frac {1}{sec(e^{x^2})}. [sec\ e^{x^2}]\)\(tan\ e^{x^2}. \frac {d}{dx} (e^{x^2})\)
\(\frac {dy}{dx}=\) \(tan\ (e^{x^2}).e^{x^2}.\frac {d}{dx}(x^2)\)
\(\frac {dy}{dx}=\) \(tan\ (e^{x^2}).e^{x^2}.2x\)
\(\frac {dy}{dx}=\) \(2x.e^{x^2}.tan\ (e^{x^2})\)

So, the correct option is (C): \(2x.e^{x^2}.tan\ (e^{x^2})\)