Q166 / 240 · KCET 2026 · Mathematics
If $y = \sqrt[3]{\tan x + y}$, then $\dfrac{dy}{dx} = $
- A$\dfrac{\tan x}{3y^2 - 1}$
- B$\dfrac{\sec^2 x}{3y - 1}$
- C$\dfrac{\tan x}{3y - 1}$
- Correct answerD$\dfrac{\sec^2 x}{3y^2 - 1}$
Solution
Given $y = \sqrt[3]{\tan x + y}$
Cubing both sides, we get:
$y^3 = \tan x + y$
Differentiating both sides with respect to $x$:
$3y^2 \dfrac{dy}{dx} = \sec^2 x + \dfrac{dy}{dx}$
Rearranging the terms to solve for $\dfrac{dy}{dx}$:
$(3y^2 - 1) \dfrac{dy}{dx} = \sec^2 x$
$\dfrac{dy}{dx} = \dfrac{\sec^2 x}{3y^2 - 1}$
Answer: $\dfrac{\sec^2 x}{3y^2 - 1}$
Cubing both sides, we get:
$y^3 = \tan x + y$
Differentiating both sides with respect to $x$:
$3y^2 \dfrac{dy}{dx} = \sec^2 x + \dfrac{dy}{dx}$
Rearranging the terms to solve for $\dfrac{dy}{dx}$:
$(3y^2 - 1) \dfrac{dy}{dx} = \sec^2 x$
$\dfrac{dy}{dx} = \dfrac{\sec^2 x}{3y^2 - 1}$
Answer: $\dfrac{\sec^2 x}{3y^2 - 1}$
Hence, the correct answer is option: 'D'.