Question

If $y=log(secx+\tan x),$ then $\frac{dy}{dx}=$.

• A. sec x$-\tan x$
• B. $\tan x$
• C. sec x
• D. $\cos x$

Answer 1

The correct option is C sec x

Given $y=log(secx+tanx)$

so, on differentiating the given function with respect to x, we get

$\Rightarrow$ $\frac{dy}{dx}$=$\frac{1}{(secx+tanx)}\cdot (secx.tanx+se{c}^2(x\left)\right)$

$\Rightarrow$ $\frac{dy}{dx}$ =$\frac{1}{(secx+tanx)}\cdot secx(tanx+secx)$

$\Rightarrow$ $\frac{dy}{dx}$ = $secx$.

so the answer of given question is $secx$.