Differentiate the following w.r.t. as indicated :
`sin^(-1)theta" w.r.t. "log(1+theta)`

Answer

To differentiate \( \sin^{-1} \theta \) with respect to \( \log(1 + \theta) \), we can use the chain rule. Let's denote:

- \( u = \sin^{-1} \theta \)
- \( v = \log(1 + \theta) \)

We need to find \( \frac{du}{dv} \).

### Step 1: Differentiate \( u \) with respect to \( \theta \)

The derivative of \( u = \sin^{-1} \theta \) is given by:

\[ \frac{du}{d\theta} = \frac{1}{\sqrt{1 - \theta^2}} \]

### Step 2: Differentiate \( v \) with respect to \( \theta \)

The derivative of \( v = \log(1 + \theta) \) is given by:

\[ \frac{dv}{d\theta} = \frac{1}{1 + \theta} \]

### Step 3: Use the chain rule to find \( \frac{du}{dv} \)

Using the chain rule, we have:

\[ \frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} \]

Substituting the derivatives we found:

\[ \frac{du}{dv} = \frac{\frac{1}{\sqrt{1 - \theta^2}}}{\frac{1}{1 + \theta}} = \frac{1 + \theta}{\sqrt{1 - \theta^2}} \]

### Final Result

Thus, the derivative of \( \sin^{-1} \theta \) with respect to \( \log(1 + \theta) \) is:

\[ \frac{du}{dv} = \frac{1 + \theta}{\sqrt{1 - \theta^2}} \]