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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If x = sec t + tan t and& y = sec t - tan t , where t is a...

If x=sect+tantx = sec t + tan t and& y=sect−tanty = sec t - tan t, where tt is a parameter, then the value of dydx\dfrac{dy}{dx} when x=13x = \dfrac{1}{\sqrt{3}} is

Options

  1. A00
  2. B−3- 3
  3. C3\sqrt{3}
  4. D13\dfrac{1}{\sqrt{3}}

Correct answer

B. −3- 3

Step-by-step solution

∵xy=1⇒y=1x∵ xy = 1 ⇒ y = \dfrac{1}{x}
dydx=−1x2\dfrac{dy}{dx} = \dfrac{- 1}{x^{2}}
at x=13,dydx=−3x = \dfrac{1}{\sqrt{3}}, \dfrac{dy}{dx} = - 3
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