Question

If sin y = sin 3t and x = sin t, then \(\frac{dy}{dx}\) =

• A. \(\frac{3}{\sqrt4-x^2}\)
• B. \(\frac{3}{\sqrt1-x^2}\)
• C. \(\frac{1}{\sqrt4-x^2}\)
• D. \(\frac{-1}{\sqrt4 - x^2}\)

Answer 1

The Correct Option is B

To solve the problem, we need to find $\frac{dy}{dx}$ given $\sin y = \sin 3t$ and $x = \sin t$.

1. Parametric Approach:
Compute $\frac{dy}{dt}$ from $\sin y = \sin 3t$, so $y = \arcsin(\sin 3t)$.
$\frac{dy}{dt} = \frac{d}{dt} \arcsin(\sin 3t) = \frac{3 \cos 3t}{\sqrt{1 - \sin^2 3t}} = \frac{3 \cos 3t}{\cos 3t} = 3$, provided $\cos 3t \neq 0$.
Compute $\frac{dx}{dt}$ from $x = \sin t$:
$\frac{dx}{dt} = \cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$.
Thus, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3}{\sqrt{1 - x^2}}$.

2. Implicit Differentiation Approach:
Since $\sin 3t = 3 \sin t - 4 \sin^3 t$ and $x = \sin t$, we have $\sin y = 3x - 4x^3$.
Differentiate with respect to $x$:
$\cos y \frac{dy}{dx} = 3 - 12x^2$.
Thus, $\frac{dy}{dx} = \frac{3 - 12x^2}{\cos y} = \frac{3(1 - 4x^2)}{\sqrt{1 - \sin^2 y}} = \frac{3(1 - 4x^2)}{\sqrt{1 - (3x - 4x^3)^2}}$.

3. Simplest Form:
The parametric approach yields $\frac{dy}{dx} = \frac{3}{\sqrt{1 - x^2}}$, which is simpler and equivalent for appropriate domains.

Final Answer:
The derivative is $\frac{3}{\sqrt{1 - x^2}}$.