If \(\sin y = x \cos(a + y)\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\cos^2(a+y)}{\cos a}\)
Since x can be written in terms of y, it is quicker to work with \(\frac{dx}{dy}\):
\( \displaystyle x = \frac{\sin y}{\cos(a+y)} \)
By the quotient rule, with \(\frac{d}{dy}\cos(a+y) = -\sin(a+y)\):
\( \displaystyle \frac{dx}{dy} = \frac{\cos y\cos(a+y) + \sin y\sin(a+y)}{\cos^2(a+y)} \)
The numerator is \(\cos\big((a+y) - y\big) = \cos a\), by the identity \(\cos(P - Q) = \cos P\cos Q + \sin P\sin Q\):
\( \displaystyle \frac{dx}{dy} = \frac{\cos a}{\cos^2(a+y)} \)
\( \displaystyle \frac{dy}{dx} = \frac{\cos^2(a+y)}{\cos a} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{\cos a}{\cos^2(a+y)}\) — Incorrect. This is \(\frac{dx}{dy}\); \(\frac{dy}{dx}\) is its reciprocal.
Option (3) \(\displaystyle\frac{\sin^2y}{\cos a}\) — Incorrect. At x = 0, y = 0 (a point on the curve) it gives 0, while \(\frac{dy}{dx} = \frac{\cos^2 a}{\cos a} = \cos a\).
Option (4) \(\displaystyle\frac{\sin^2y}{\cos (a+y)}\) — Incorrect. At x = 0, y = 0 it also gives 0 instead of \(\cos a\).