' find dy/dx if x=3sin t-2 sin ^3t, y=3 cos t -2cos^3t`

Answer

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

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

We start with the equation for \(x\):
\[ x = 3\sin t - 2\sin 3t \]
Now, we differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = 3\cos t - 2\frac{d}{dt}(\sin 3t) \]
Using the chain rule, we find:
\[ \frac{d}{dt}(\sin 3t) = 3\cos 3t \]
Thus, substituting this back, we have:
\[ \frac{dx}{dt} = 3\cos t - 2(3\cos 3t) = 3\cos t - 6\cos 3t \]

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

Next, we differentiate \(y\):
\[ y = 3\cos t - 2\cos 3t \]
Differentiating \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = -3\sin t + 2\frac{d}{dt}(\cos 3t) \]
Using the chain rule again:
\[ \frac{d}{dt}(\cos 3t) = -3\sin 3t \]
So, substituting this back, we have:
\[ \frac{dy}{dt} = -3\sin t + 2(-3\sin 3t) = -3\sin t - 6\sin 3t \]

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

Now, we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = \frac{-3\sin t - 6\sin 3t}{3\cos t - 6\cos 3t} \]

### Step 4: Simplify the expression

We can simplify the expression if possible, but in this case, it is already in a simplified form. Thus, we have:
\[ \frac{dy}{dx} = \frac{-3\sin t - 6\sin 3t}{3\cos t - 6\cos 3t} \]

### Final Answer

So, the final answer for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{-3\sin t - 6\sin 3t}{3\cos t - 6\cos 3t} \]

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