Question

Y=e^x+cot x-tan x (differentiation)​

Answer 1

$y = {e}^{x} + cotx - tanx \\ \frac{dy}{dx} = \frac{d}{dx} {e}^{x} + \frac{d}{dx} cotx - \frac{d}{dx} tanx \\ \frac{dy}{dx} = {e}^{x} - {cosec}^{2} x - {sec}^{2} x$