`x=3sin t-2 sin ^3t, y=3 cos t -2cos^3t`then find `dy/dx`

Answer

To find \(\frac{dy}{dx}\) for the given parametric equations \(x = 3\sin t - 2\sin^3 t\) and \(y = 3\cos t - 2\cos^3 t\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(t\)

We start by differentiating \(x\) with respect to \(t\):

\[ x = 3\sin t - 2\sin^3 t \]

Using the chain rule, we differentiate:

\[ \frac{dx}{dt} = 3\cos t - 2 \cdot 3\sin^2 t \cdot \frac{d}{dt}(\sin t) \]

This simplifies to:

\[ \frac{dx}{dt} = 3\cos t - 6\sin^2 t \cos t \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Next, we differentiate \(y\):

\[ y = 3\cos t - 2\cos^3 t \]

Again, using the chain rule:

\[ \frac{dy}{dt} = -3\sin t + 2 \cdot 3\cos^2 t \cdot (-\sin t) \]

This simplifies to:

\[ \frac{dy}{dt} = -3\sin t + 6\cos^2 t \sin t \]

### Step 3: Find \(\frac{dy}{dx}\)

Now, we can find \(\frac{dy}{dx}\) using the formula:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = \frac{-3\sin t + 6\cos^2 t \sin t}{3\cos t - 6\sin^2 t \cos t} \]

### Step 4: Simplify the expression

We can factor out \(\sin t\) from the numerator and \(\cos t\) from the denominator:

\[ \frac{dy}{dx} = \frac{\sin t(-3 + 6\cos^2 t)}{\cos t(3 - 6\sin^2 t)} \]

This can be further simplified:

\[ \frac{dy}{dx} = \frac{\sin t(6\cos^2 t - 3)}{\cos t(3 - 6\sin^2 t)} \]

Using the identity \(6\cos^2 t - 3 = 3(2\cos^2 t - 1) = 3\cos(2t)\) and \(3 - 6\sin^2 t = 3(1 - 2\sin^2 t) = 3\cos(2t)\):

\[ \frac{dy}{dx} = \frac{\sin t \cdot 3\cos(2t)}{\cos t \cdot 3\cos(2t)} \]

The \(3\cos(2t)\) cancels out (assuming \(\cos(2t) \neq 0\)):

\[ \frac{dy}{dx} = \frac{\sin t}{\cos t} = \tan t \]

### Final Result

Thus, we have:

\[ \frac{dy}{dx} = \tan t \]