If \( y=\tan^{-1}(x+y) \), then \( \displaystyle\frac{d^2y}{dx^2} \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{-2(1+(x+y)^2)}{(x+y)^5} \)
Let u = x + y. Differentiating \( y=\tan^{-1}u \) implicitly,
\( \displaystyle \frac{dy}{dx}=\frac{1}{1+u^2}\left(1+\frac{dy}{dx}\right) \)
\( \displaystyle (1+u^2)\frac{dy}{dx}=1+\frac{dy}{dx} \)
\( \displaystyle u^2\frac{dy}{dx}=1 \)
\( \displaystyle \frac{dy}{dx}=\frac{1}{(x+y)^2} \)
Differentiating again; by the chain rule, \( \frac{d}{dx}(x+y)=1+\frac{dy}{dx} \), not 1:
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{2}{(x+y)^3}\left(1+\frac{dy}{dx}\right) \)
\( \displaystyle =-\frac{2}{(x+y)^3}\left(1+\frac{1}{(x+y)^2}\right) \)
\( \displaystyle =-\frac{2}{(x+y)^3}\cdot\frac{(x+y)^2+1}{(x+y)^2} \)
\( \displaystyle =\frac{-2(1+(x+y)^2)}{(x+y)^5} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{-2(1+(x+y)^2)}{(x+y)^3} \) — Incorrect. The power of \( (x+y) \) in the denominator is 5, not 3.
Option (2) \( \displaystyle\frac{-(1+(x+y)^2)}{(x+y)^3} \) — Incorrect. The numerator has the factor 2, and the power of \( (x+y) \) in the denominator is 5, not 3.
Option (4) \( \displaystyle\frac{-(1+(x+y)^2)}{(x+y)^5} \) — Incorrect. The factor 2 from differentiating \( (x+y)^{-2} \) is missing.