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Question: If x + y = t – \(\frac{1}{t}\), x<sup>2</sup> + y<sup>2</sup> = t<sup>2</sup> + \(\frac{1}{t^{2}}\),...

If x + y = t – 1t\frac{1}{t}, x2 + y2 = t2 + 1t2\frac{1}{t^{2}}, then dydx\frac{dy}{dx} is equal to

A

1x\frac{1}{x}

B

−1x- \frac{1}{x}

C

1x2\frac{1}{x^{2}}

D

−1x2- \frac{1}{x^{2}}

Answer

1x2\frac{1}{x^{2}}

Explanation

Solution

x + y = t – 1t\frac{1}{t}, x2 + y2 = t2 + 1t2\frac{1}{t^{2}}

(x + y)2 = t2 +1t2\frac{1}{t^{2}} – 2

⇒ x2 + y2 + 2xy = x2 + y2 – 2

⇒ xy = – 1 ⇒ y = −1x\frac{- 1}{x} ⇒ y ′ = 1x2\frac{1}{x^{2}}