Question

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

• A. $\frac{\sin \frac{a}{2}}{\sin(a + y)}$
• B. $\frac{\sin(a + y)}{\sin^2 a}$
• C. $\frac{\sin(a + y)}{\sin a}$
• D. $\frac{\sin^2(a + y)}{\sin a}$

Answer 1

Answer: (D)

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

Answer 2

The given equation is:

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

Differentiate both sides with respect to $x$:

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

Rearrange to isolate $\frac{dy}{dx}$:

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

Simplify:

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

From the original equation $\sin y = x \sin(a + y)$, rewrite $x$ as:

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

Substitute $x$ into the denominator:

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

Simplify the denominator:

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

Substitute this back into $\frac{dy}{dx}$:

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

Thus:

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