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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If x = 1 - t^ 2 1 + t^ 2 and y = 2t 1 + t^ 2 , then(dy)/(dx) is...

If x=1−t21+t2x = \dfrac{1 - t^{2}}{1 + t^{2}} and y=2t1+t2,thendydxy = \dfrac{2t}{1 + t^{2}}, then\dfrac{dy}{dx} is equal to

Options

  1. A−yx- \dfrac{y}{x}
  2. Byx\dfrac{y}{x}
  3. C−xy- \dfrac{x}{y}
  4. Dxy\dfrac{x}{y}

Correct answer

C. −xy- \dfrac{x}{y}

Step-by-step solution

∵x=1−t21+t2andy=2t1+t2∵ x = \dfrac{1 - t^{2}}{1 + t^{2}}andy = \dfrac{2t}{1 + t^{2}}

Put t=tanθt = tan \theta in both the equations, we get

x=1−tan⁡2θ1+tan⁡2θ=cos⁡2θx = \dfrac{1 - \tan^{2} \theta}{1 + \tan^{2} \theta} = \cos 2\theta ...(i)

and y=2tanθ1+tan⁡2θ=sin⁡2θy = \dfrac{2tan \theta}{1 + \tan^{2} \theta} = \sin 2\theta ...(ii)

On differentiating both the eqs.(i) and (ii), we get

dxdθ=−2sin⁡2θanddydθ=2cos⁡2θ\dfrac{dx}{d\theta} = - 2\sin 2\theta and\dfrac{dy}{d\theta} = 2\cos 2\theta

Therefore, dydx=dydθdxdθ=−cos⁡2θsin⁡2θ=−xy\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = - \dfrac{\cos 2\theta}{\sin 2\theta} = - \dfrac{x}{y}
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