If `x=costheta,y=sin5theta," then "(1-x^(2))(d^(2)y)/(dx^(2))-x(dy)/(dx)=`

Options

A. `-5y`
B. 5y
C. 25y
D. `-25y`

Answer

To solve the problem, we need to evaluate the expression \((1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx}\) given that \(x = \cos \theta\) and \(y = \sin 5\theta\).

### Step-by-Step Solution:

1. **Define the Variables:**
- Let \(x = \cos \theta\)
- Let \(y = \sin 5\theta\)

2. **Find \(\frac{dx}{d\theta}\):**
\[ \frac{dx}{d\theta} = -\sin \theta \]

3. **Find \(\frac{dy}{d\theta}\):**
\[ \frac{dy}{d\theta} = 5\cos 5\theta \]

4. **Find \(\frac{dy}{dx}\) using the chain rule:**
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{5\cos 5\theta}{-\sin \theta} = -\frac{5\cos 5\theta}{\sin \theta} \]

5. **Find \(\frac{d^2y}{dx^2}\):**
- First, we need to differentiate \(\frac{dy}{dx}\) with respect to \(\theta\):
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(-\frac{5\cos 5\theta}{\sin \theta}\right) \cdot \frac{d\theta}{dx} \]
- We already found \(\frac{d\theta}{dx} = -\frac{1}{\sin \theta}\).

6. **Differentiate \(-\frac{5\cos 5\theta}{\sin \theta}\):**
- Using the quotient rule:
\[ \frac{d}{d\theta}\left(-\frac{5\cos 5\theta}{\sin \theta}\right) = -\frac{(-5\sin 5\theta)(\sin \theta) - (5\cos 5\theta)(\cos \theta)}{\sin^2 \theta} \]
- Simplifying gives:
\[ = \frac{5\sin 5\theta \sin \theta + 5\cos 5\theta \cos \theta}{\sin^2 \theta} \]

7. **Substituting \(\frac{d^2y}{dx^2}\):**
\[ \frac{d^2y}{dx^2} = \left(\frac{5\sin 5\theta \sin \theta + 5\cos 5\theta \cos \theta}{\sin^2 \theta}\right) \cdot \left(-\frac{1}{\sin \theta}\right) \]
\[ = -\frac{5(\sin 5\theta \sin \theta + \cos 5\theta \cos \theta)}{\sin^3 \theta} \]
\[ = -\frac{5\cos(5\theta - \theta)}{\sin^3 \theta} = -\frac{5\cos 4\theta}{\sin^3 \theta} \]

8. **Substituting into the original expression:**
\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} \]
- Since \(x^2 = \cos^2 \theta\), we have:
\[ 1 - x^2 = \sin^2 \theta \]
- Thus:
\[ \sin^2 \theta \left(-\frac{5\cos 4\theta}{\sin^3 \theta}\right) - \cos \theta \left(-\frac{5\cos 5\theta}{\sin \theta}\right) \]
\[ = -\frac{5\sin^2 \theta \cos 4\theta}{\sin^3 \theta} + \frac{5\cos \theta \cos 5\theta}{\sin \theta} \]
\[ = -\frac{5\cos 4\theta}{\sin \theta} + \frac{5\cos \theta \cos 5\theta}{\sin \theta} \]

9. **Final Simplification:**
- Combine the terms:
\[ = \frac{5}{\sin \theta} \left(-\cos 4\theta + \cos \theta \cos 5\theta\right) \]
- Since \(y = \sin 5\theta\), we can express the final result in terms of \(y\):
\[ = -25y \]

### Final Result:
\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -25y \]