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Question: If \(x = \exp\left\{ \tan^{- 1}\left( \frac{y - x^{2}}{x^{2}} \right) \right\}\)then \(\frac{dy}{dx}...

If x=exp⁡{tan⁡−1(y−x2x2)}x = \exp\left\{ \tan^{- 1}\left( \frac{y - x^{2}}{x^{2}} \right) \right\}then dydx\frac{dy}{dx}equals

A

2x[1+tan⁡(log⁡x)]+xsec⁡2(log⁡x)2x\lbrack 1 + \tan(\log x)\rbrack + x\sec^{2}(\log x)

B

x[1+tan⁡(log⁡x)]+sec⁡2(log⁡x)x\lbrack 1 + \tan(\log x)\rbrack + \sec^{2}(\log x)

C

2x[1+tan⁡(log⁡x)]+x2sec⁡2(log⁡x)2x\lbrack 1 + \tan(\log x)\rbrack + x^{2}\sec^{2}(\log x)

D

2x[1+tan⁡(log⁡x)]+sec⁡2(log⁡x)2x\lbrack 1 + \tan(\log x)\rbrack + \sec^{2}(\log x)

Answer

2x[1+tan⁡(log⁡x)]+xsec⁡2(log⁡x)2x\lbrack 1 + \tan(\log x)\rbrack + x\sec^{2}(\log x)

Explanation

Solution

x=exp⁡{tan⁡−1(y−x2x2)}x = \exp\left\{ \tan^{- 1}\left( \frac{y - x^{2}}{x^{2}} \right) \right\} ⇒ log⁡x=tan⁡−1(y−x2x2)\log x = \tan^{- 1}\left( \frac{y - x^{2}}{x^{2}} \right)

⇒ y−x2x2=tan⁡(log⁡x)\frac{y - x^{2}}{x^{2}} = \tan(\log x) ⇒ y=x2tan⁡(log⁡x)+x2y = x^{2}\tan(\log x) + x^{2}

⇒ dydx=2x.tan⁡(log⁡x)+x2.sec⁡2(log⁡x)x+2x\frac{dy}{dx} = 2x.\tan(\log x) + x^{2}.\frac{\sec^{2}(\log x)}{x} + 2x

⇒ dydx=2xtan⁡(log⁡x)+xsec⁡2(log⁡x)+2x\frac{dy}{dx} = 2x\tan(\log x) + x\sec^{2}(\log x) + 2x

⇒ dydx=2x[1+tan⁡(log⁡x)]+xsec⁡2(log⁡x)\frac{dy}{dx} = 2x\lbrack 1 + \tan(\log x)\rbrack + x\sec^{2}(\log x)