If `y=log(1+ theta), x= sin^(-1) theta`, then `(dy)/(dx)`=

Options

A. `sqrt(1/(1+theta))`
B. `(1+ theta)/(1- theta)`
C. `sqrt((1-theta)/(1+ theta))`
D. `sqrt((1+ theta)/(1- theta))`

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given that \(y = \log(1 + \theta)\) and \(x = \sin^{-1}(\theta)\). We will use the chain rule for differentiation.

### Step-by-Step Solution:

1. **Identify the functions**:
- \(y = \log(1 + \theta)\)
- \(x = \sin^{-1}(\theta)\)

2. **Differentiate \(y\) with respect to \(\theta\)**:
\[ \frac{dy}{d\theta} = \frac{d}{d\theta} \log(1 + \theta) = \frac{1}{1 + \theta} \cdot \frac{d}{d\theta}(1 + \theta) = \frac{1}{1 + \theta} \]

3. **Differentiate \(x\) with respect to \(\theta\)**:
\[ \frac{dx}{d\theta} = \frac{d}{d\theta} \sin^{-1}(\theta) = \frac{1}{\sqrt{1 - \theta^2}} \]

4. **Apply the chain rule**:
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} = \frac{dy}{d\theta} \cdot \frac{1}{\frac{dx}{d\theta}} \]

5. **Substitute the derivatives**:
\[ \frac{dy}{dx} = \frac{1}{1 + \theta} \cdot \frac{1}{\frac{1}{\sqrt{1 - \theta^2}}} = \frac{1}{1 + \theta} \cdot \sqrt{1 - \theta^2} \]

6. **Final expression**:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - \theta^2}}{1 + \theta} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - \theta^2}}{1 + \theta} \]