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

Answer

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

- \( U = \cos^{-1}(\theta) \)
- \( V = \log(1+\theta) \)

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

The derivative of \( U \) with respect to \( \theta \) is given by:

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

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

The derivative of \( V \) with respect to \( \theta \) 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 can express \( \frac{dU}{dV} \) as:

\[ \frac{dU}{dV} = \frac{dU/d\theta}{dV/d\theta} \]

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 \( \cos^{-1}(\theta) \) with respect to \( \log(1+\theta) \) is:

\[ \frac{dU}{dV} = -\frac{1 + \theta}{\sqrt{1 - \theta^2}} \]