Question

Differentiate and find $\frac{dy}{dx}$ :

$\tan x+y=\sec x-3x^2$

Answer 1

$tanx+y=secx-3x^2$

$se{c}^{2}x+\frac{dy}{dx}=secxtanx-6x$

$\frac{dy}{dx}=secxtanx-6x-se{c}^{2}x$