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

Answer: (C)

$2x.e^{x^2}.tan\ e^{x^2}$

Answer 2

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): $2x.e^{x^2}.tan\ (e^{x^2})$