KcetDifferentiationKCET 2026MCQ+1 / -02026If $y = \sqrt[3]{\tan x + y}$, then $\dfrac{dy}{dx} = $A$\dfrac{\tan x}{3y^2 - 1}$B$\dfrac{\sec^2 x}{3y - 1}$C$\dfrac{\tan x}{3y - 1}$D$\dfrac{\sec^2 x}{3y^2 - 1}$Check AnswerClear SelectionReveal AnswerShow Explanation
