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

NTA Abhyas 2020 · Mathematics · Differentiation

If y = ta n ^ - 1 ( x 1 + 6 x ^ 2 ) + ta n ^ - 1 ((2x - 1)/(2x +...

If y=tan−1(x1+6x2)+tan−1(2x−12x+1)(∀x>0)y = ta{n}^{- 1}\left(\dfrac{x}{1 + 6{x}^{2}}\right) + ta{n}^{- 1}\left(\dfrac{2x - 1}{2x + 1}\right)\left(\forall x > 0\right), then dydx\dfrac{dy}{dx} is equal to

Options

  1. A31+9x2\dfrac{3}{1 + 9{x}^{2}}
  2. B11+6x2\dfrac{1}{1 + 6{x}^{2}}
  3. C11+6x2+11+x2\dfrac{1}{1 + 6{x}^{2}} + \dfrac{1}{1 + {x}^{2}}
  4. D31+6x2\dfrac{3}{1 + 6{x}^{2}}

Correct answer

A. 31+9x2\dfrac{3}{1 + 9{x}^{2}}

Step-by-step solution

y=tan−1(3x−2x1+3x⋅2x)+tan−1(2x−11+2x⋅1)y = ta{n}^{- 1}\left(\dfrac{3x - 2x}{1 + 3x ⋅ 2x}\right) + ta{n}^{- 1}\left(\dfrac{2x - 1}{1 + 2x ⋅ 1}\right)
y=tan−13x−tan−12x+tan−12x−tan⁡−11y = ta{n}^{- 1}3x - ta{n}^{- 1}2x + ta{n}^{- 1}2x - \tan^{- 1}1
y=tan−13x−π4⇒dydx=31+9x2y = ta{n}^{- 1}3x - \dfrac{\pi}{4} ⇒ \dfrac{dy}{dx} = \dfrac{3}{1 + 9{x}^{2}}
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