Find `(d^(2)y)/(dx^(2))` in the following
`x=acos^(3)theta,y=bsin^(3)theta`

Answer

To find the second derivative \(\frac{d^2y}{dx^2}\) given the parametric equations \(x = a \cos^3 \theta\) and \(y = b \sin^3 \theta\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(\theta\)
Given:
\[ x = a \cos^3 \theta \]
We differentiate \(x\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)
Given:
\[ y = b \sin^3 \theta \]
We differentiate \(y\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = b \cdot 3 \sin^2 \theta \cdot \cos \theta = 3b \sin^2 \theta \cos \theta \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3b \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]
Simplifying this:
\[ \frac{dy}{dx} = -\frac{b \sin \theta}{a \cos \theta} = -\frac{b}{a} \tan \theta \]

### Step 4: Differentiate \(\frac{dy}{dx}\) with respect to \(\theta\)
Now we need to differentiate \(\frac{dy}{dx}\) with respect to \(\theta\):
\[ \frac{d}{d\theta}\left(\frac{dy}{dx}\right) = -\frac{b}{a} \sec^2 \theta \]

### Step 5: Find \(\frac{d^2y}{dx^2}\)
Using the formula for the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{d\theta}} \]
Substituting the values we found:
\[ \frac{d^2y}{dx^2} = \left(-\frac{b}{a} \sec^2 \theta\right) \cdot \frac{1}{-3a \cos^2 \theta \sin \theta} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{b \sec^2 \theta}{3a^2 \cos^2 \theta \sin \theta} \]
Since \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\), we can further simplify:
\[ \frac{d^2y}{dx^2} = \frac{b}{3a^2 \sin \theta \cos^4 \theta} \]

### Final Result
Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = \frac{b}{3a^2 \sin \theta \cos^4 \theta} \]