If `y=sinx + tan^(-1)(1)`, find dy/dx

Answer

To find the derivative of the function \( y = \sin x + \tan^{-1}(1) \) with respect to \( x \), we can follow these steps:

### Step 1: Identify the components of the function
We have:
\[ y = \sin x + \tan^{-1}(1) \]

### Step 2: Simplify the constant
We know that:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]
Thus, we can rewrite \( y \) as:
\[ y = \sin x + \frac{\pi}{4} \]

### Step 3: Differentiate both sides with respect to \( x \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sin x) + \frac{d}{dx}\left(\frac{\pi}{4}\right) \]

### Step 4: Apply the differentiation rules
Using the differentiation rules:
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of a constant (in this case, \( \frac{\pi}{4} \)) is \( 0 \).

Thus, we have:
\[ \frac{dy}{dx} = \cos x + 0 \]

### Step 5: Write the final answer
Therefore, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \cos x \]

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