If y=tan^(-1)x/(1+6x^2)+tan^(-1)(2x-1)/(2x+1) ∀x>0, then dy/dx is equal to

Correct answer: (A) 3/(1+9x^2). y=tan^(-1)(3x-2x)/(1+3x⋅2x)+tan^(-1)(2x-1)/(1+2x⋅1) y=tan^(-1)3x-tan^(-1)2x+tan^(-1)2x-tan^(-1)1 y=tan^(-1)3x-(π)/4⇒dy/dx=3/(1+9x^2)

Q24 / 46 · NTA Abhyas (JEE Main) 2020 · Mathematics · Differentiation

If $y = t a n^{- 1} (\frac{x}{1 + 6 x^{2}}) + t a n^{- 1} (\frac{2 x - 1}{2 x + 1}) (\forall x > 0)$ , then $\frac{d y}{d x}$ is equal to

Correct answer
A
$\frac{3}{1 + 9 x^{2}}$
B
$\frac{1}{1 + 6 x^{2}}$
C
$\frac{1}{1 + 6 x^{2}} + \frac{1}{1 + x^{2}}$
D
$\frac{3}{1 + 6 x^{2}}$

Solution

y = t a n^{- 1} (\frac{3 x - 2 x}{1 + 3 x \cdot 2 x}) + t a n^{- 1} (\frac{2 x - 1}{1 + 2 x \cdot 1})

y = t a n^{- 1} 3 x - t a n^{- 1} 2 x + t a n^{- 1} 2 x - \tan^{- 1} 1

y = t a n^{- 1} 3 x - \frac{π}{4} \Rightarrow \frac{d y}{d x} = \frac{3}{1 + 9 x^{2}}

Hence, the correct answer is option: 'A'.

If y=tan-1x1+6x2+tan-12x-12x+1 ∀x>0, then dydx is equal to

Solution

More Differentiation Questions

If $y = t a n^{- 1} (\frac{x}{1 + 6 x^{2}}) + t a n^{- 1} (\frac{2 x - 1}{2 x + 1}) (\forall x > 0)$ , then $\frac{d y}{d x}$ is equal to