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

Explanation

To differentiate the function \( y = \cos(\tan(\sqrt{x+1})) \), we will use the chain rule. The chain rule states that if you have a composite function, the derivative can be found by multiplying the derivative of the outer function by the derivative of the inner function.

### Step-by-Step Solution:

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

2. **Differentiate the outer function**:
- The derivative of \( y = \cos(u) \) is \( \frac{dy}{du} = -\sin(u) \).

3. **Differentiate the first inner function**:
- The derivative of \( u = \tan(v) \) is \( \frac{du}{dv} = \sec^2(v) \).

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

5. **Apply the chain rule**:
- By the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = -\sin(\tan(\sqrt{x+1})) \cdot \sec^2(\sqrt{x+1}) \cdot \frac{1}{2\sqrt{x+1}} \]

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