Question

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

(A) $\frac{1}{2} \sec ^2 x$

(B) $\sec ^2 x$

(C) $\sec x \tan x$

(D) $e^{\frac{1}{2} \log \left(1+\tan ^2 x\right)}$

Answer 1

$y=e^{\frac{1}{2} \log \left(a+\tan ^2 x\right)} \Rightarrow y=\left(\sec ^2 x\right)^{1 / 2}=\sec x$

$\therefore \quad \frac{d y}{d x}=\sec x \tan x$

So, The correct option is (C).