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

Explanation

To differentiate the function \( y = \cos(\tan(\sqrt{x+1})) \), we will apply the chain rule of differentiation step by step.

### Step 1: Identify the outer and inner functions
Let:
- \( u = \tan(\sqrt{x+1}) \)
- \( y = \cos(u) \)

### Step 2: Differentiate the outer function
Using the chain rule, we differentiate \( y \) with respect to \( u \):
\[ \frac{dy}{du} = -\sin(u) \]

### Step 3: Differentiate the inner function
Next, we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = \sec^2(\sqrt{x+1}) \cdot \frac{d}{dx}(\sqrt{x+1}) \]

### Step 4: Differentiate \( \sqrt{x+1} \)
Now we differentiate \( \sqrt{x+1} \):
\[ \frac{d}{dx}(\sqrt{x+1}) = \frac{1}{2\sqrt{x+1}} \]

### Step 5: Combine the derivatives
Now we can substitute back into our expression for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \sec^2(\sqrt{x+1}) \cdot \frac{1}{2\sqrt{x+1}} \]

### Step 6: Apply the chain rule
Now we can combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin(\tan(\sqrt{x+1})) \cdot \left(\sec^2(\sqrt{x+1}) \cdot \frac{1}{2\sqrt{x+1}}\right) \]

### Final Result
Thus, the derivative of \( y = \cos(\tan(\sqrt{x+1})) \) is:
\[ \frac{dy}{dx} = -\frac{\sin(\tan(\sqrt{x+1})) \cdot \sec^2(\sqrt{x+1})}{2\sqrt{x+1}} \]