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

The correct option is C ysecx
Given, $y=secx+tanx$

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

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

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

$\Rightarrow$ $\frac{dy}{dx}$ = $ysecx$

so the answer of given question is $ysecx$.