Question:medium

\(\text{ If } \sin y = x \sin(a + y), \text{ then } \frac{dy}{dx} \text{ is:}\)

Show Hint

When differentiating an implicit equation, always ensure to apply the chain rule correctly. In cases where the equation involves trigonometric functions, be sure to use their derivative identities. Additionally, when isolating \(\frac{dy}{dx}\), carefully rearrange terms to simplify the expression. In this problem, substituting the expression for \(x\) into the denominator helped simplify the final result.

Updated On: Mar 27, 2026
  • \( \frac{\sin \frac{a}{2}}{\sin(a + y)} \)
  • \( \frac{\sin(a + y)}{\sin^2 a} \)
  • \( \frac{\sin(a + y)}{\sin a} \)
  • \( \frac{\sin^2(a + y)}{\sin a} \)
Show Solution

The Correct Option is D

Solution and Explanation

The provided equation is:

\(\sin y = x \sin(a + y)\).

Differentiating both sides with respect to \(x\) yields:

\(\cos y \frac{dy}{dx} = \sin(a + y) + x \cos(a + y) \frac{dy}{dx}\).

Rearranging to isolate \(\frac{dy}{dx}\) gives:

\(\frac{dy}{dx} (\cos y - x \cos(a + y)) = \sin(a + y)\).

Simplifying the expression for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\sin(a + y)}{\cos y - x \cos(a + y)}\).

From the initial equation \(\sin y = x \sin(a + y)\), we express \(x\) as:

\(x = \frac{\sin y}{\sin(a + y)}\).

Substituting this expression for \(x\) into the denominator:

\(\cos y - x \cos(a + y) = \cos y - \frac{\sin y \cos(a + y)}{\sin(a + y)}\).

Simplifying the denominator further:

\(\cos y - x \cos(a + y) = \frac{\cos y \sin(a + y) - \sin y \cos(a + y)}{\sin(a + y)} = \frac{\sin a}{\sin(a + y)}\).

Substituting this simplified denominator back into the expression for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\sin(a + y)}{\frac{\sin a}{\sin(a + y)}} = \frac{\sin^2(a + y)}{\sin a}\).

Therefore, the final result is:

\(\boxed{\frac{\sin^2(a + y)}{\sin a}}\).

Was this answer helpful?
0


Questions Asked in CUET (UG) exam