If \(\tan(x + y) = x\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{x^{2}}{1 + x^{2}}\)
By implicit differentiation with respect to x:
\( \displaystyle \sec^{2}(x + y)\left(1 + \frac{dy}{dx}\right) = 1 \)
Using \(\sec^{2}\theta = 1 + \tan^{2}\theta\) and \(\tan(x + y) = x\):
\(\sec^{2}(x + y) = 1 + x^{2}\)
\( \displaystyle 1 + \frac{dy}{dx} = \frac{1}{1 + x^{2}} \)
\( \displaystyle \frac{dy}{dx} = \frac{1}{1 + x^{2}} - 1 = -\frac{x^{2}}{1 + x^{2}} \)