Question

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

• A. $\frac{1}{2}se{c}^{2}x$
• B. $se{c}^{2}x$
• C. $secxtanx$
• D. $e^{\frac{1}{2}log\left(1+ta{n}^{2}x\right)}$

Answer 1

Answer: (C)

$y=e^{\frac{1}{2}log\left(1+ta{n}^{2}x\right)}$
$={\left(se{c}^{2}x\right)}^{1/2}=secx$
$\therefore \frac{dy}{dx}=secxtanx$