If $x = e^{\tan^{-1}\frac{(y - x^2)}{x^2}}$, then $\frac{dy}{dx}$ equals
- A. $2x[1 + \tan(\log x)] + x \sec^2(\log x)$
- B. $x[1 + \tan(\log x)] + \sec^2(\log x)$
- C. $2x[1 + \tan(\log x)] + x^2 \sec^2(\log x)$
- D. $2x[1 + \tan(\log x)] + \sec^2(\log x)$
✅ Correct Answer: $2x[1 + \tan(\log x)] + x \sec^2(\log x)$
Explanation
Differentiate implicitly using the chain rule and logarithmic differentiation. Simplifying gives $\frac{dy}{dx} = 2x[1 + \tan(\log x)] + x\sec^2(\log x)$.
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