If y=[3]{ x+y}, then d y/d x=

Correct Answer is : sec⁡2x3y2−1{ ² x}{3 y²-1}3y2−1sec2x

Correct Answer is : sec⁡2x3y2−1{ ² x}{3 y²-1}3y2−1sec2x

Test Index

KCET 2026 Mathematics Solved Paper

Question : 35 of 60

Marks: +1, -0

y=\sqrt[3]{\tan x+y}

\frac{d y}{d x}=

Solution:

Concept:

Differentiate an implicit function after eliminating the cube root.

Explanation:

Given:

y = \sqrt[3]{\tan x + y}

Cube both sides to remove the cube root:

y^3 = \tan x + y

Differentiate both sides with respect to

x

Using the chain rule on

y^3

\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}

Differentiate the right side:

\frac{d}{dx}(\tan x + y) = \sec^2 x + \frac{dy}{dx}

Thus:

3y^2 \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}

Rearrange the terms:

3y^2 \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x

Factor out

\frac{dy}{dx}

(3y^2 - 1)\frac{dy}{dx} = \sec^2 x

Therefore:

\frac{dy}{dx} = \frac{\sec^2 x}{3y^2 - 1}

Answer:

\frac{dy}{dx} = \frac{\sec^2 x}{3y^2 - 1}

Hence, the correct option is D.

Go to Question:

Prev Question

Next Question