Question

If $y=e^{\frac{1}{2}log\left(1+tan^{2}\,x\right)}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{2}sec^{2}x$
• B. $sec^{2}x$
• C. $secx\,tanx$
• D. $e^{\frac{1}{2}log\left(1+tan^{2}\,x\right)}$

Answer 1

Answer: (C)

$secx\,tanx$

Answer 2

$y=e^{\frac{1}{2}log\left(1+tan^{2}\,x\right)}$ $=\left(sec^{2}\,x\right)^{1/2}=sec\,x$ $\therefore \frac{dy}{dx}=sec\,x\,tan\,x$