Skip to main contentSkip to solution

If y=tan⁡−1(x+y)y=\tan^{-1}(x+y), then d2ydx2\frac{d^2y}{dx^2} is equal to:

Solution

✅ Correct Option: 3

From tan⁡y=x+y\tan y=x+y: sec⁡2y y′=1+y′\sec^2y\,y'=1+y', so y′tan⁡2y=1y'\tan^2y=1 and y′=1(x+y)2y'=\frac{1}{(x+y)^2}.

Differentiating, y′′=−2(x+y)−3(1+y′)=−2(x+y)−3⋅(x+y)2+1(x+y)2y''=-2(x+y)^{-3}(1+y')=-2(x+y)^{-3}\cdot\frac{(x+y)^2+1}{(x+y)^2}

=−2(1+(x+y)2)(x+y)5=\frac{-2\left(1+(x+y)^2\right)}{(x+y)^5}.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question