TS EAMCET · Maths · Differentiation
If \(x=\sqrt{1-\tan y}\), then \(\frac{d y}{d x}=\)
- A \(\frac{2 x}{x^4+2 x^2+2}\)
- B \(-\frac{2 x}{x^4-2 x^2+2}\)
- C \(\frac{2 x}{x^4-2 x^2+2}\)
- D \(-\frac{2 x}{x^4+2 x^2+2}\)
Answer & Solution
Correct Answer
(B) \(-\frac{2 x}{x^4-2 x^2+2}\)
Step-by-step Solution
Detailed explanation
\(\tan y = 1 - x^2\) \(\sec^2 y \frac{dy}{dx} = -2x\) \((1 + \tan^2 y) \frac{dy}{dx} = -2x\) \((1 + (1 - x^2)^2) \frac{dy}{dx} = -2x\) \((1 + 1 - 2x^2 + x^4) \frac{dy}{dx} = -2x\) \((x^4 - 2x^2 + 2) \frac{dy}{dx} = -2x\) \(\frac{dy}{dx} = -\frac{2x}{x^4 - 2x^2 + 2}\)
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