Differentiate `cos(tansqrt(x+1))`

Explanation

To differentiate the function \( y = \cos(\tan(\sqrt{x + 1})) \), we will apply the chain rule repeatedly. Let's go through the steps:

### Step-by-Step Solution:

1. **Identify the outer function and inner functions**:
- The outer function is \( \cos(u) \) where \( u = \tan(v) \) and \( v = \sqrt{x + 1} \).

2. **Differentiate the outer function**:
- The derivative of \( \cos(u) \) is \( -\sin(u) \). Therefore, we have:
\[ \frac{dy}{dx} = -\sin(\tan(\sqrt{x + 1})) \cdot \frac{du}{dx} \]

3. **Differentiate the middle function**:
- Now, differentiate \( u = \tan(v) \). The derivative of \( \tan(v) \) is \( \sec^2(v) \). Thus:
\[ \frac{du}{dx} = \sec^2(\sqrt{x + 1}) \cdot \frac{dv}{dx} \]

4. **Differentiate the innermost function**:
- Now, differentiate \( v = \sqrt{x + 1} \). The derivative of \( \sqrt{x + 1} \) is:
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{x + 1}} \]

5. **Combine all parts**:
- Substitute \( \frac{dv}{dx} \) back into \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \sec^2(\sqrt{x + 1}) \cdot \frac{1}{2\sqrt{x + 1}} \]
- Now, substitute \( \frac{du}{dx} \) back into \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\sin(\tan(\sqrt{x + 1})) \cdot \left( \sec^2(\sqrt{x + 1}) \cdot \frac{1}{2\sqrt{x + 1}} \right) \]

6. **Final expression**:
- Therefore, the derivative is:
\[ \frac{dy}{dx} = -\frac{\sin(\tan(\sqrt{x + 1})) \cdot \sec^2(\sqrt{x + 1})}{2\sqrt{x + 1}} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{\sin(\tan(\sqrt{x + 1})) \cdot \sec^2(\sqrt{x + 1})}{2\sqrt{x + 1}} \]