KCET · Maths · Differentiation
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 \(\dfrac{\tan x}{3y - 1}\)
- D \(\dfrac{\sec^2 x}{3y^2 - 1}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{\sec^2 x}{3y^2 - 1}\)
Step-by-step Solution
Detailed explanation
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: \(\dfrac{\sec^2 x}{3y^2 - 1}\)
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