If the derivative of `tan^(-1) (a + bx)` takes the value 1 at x= 0, the `1 +a^(2)=`

Options

A. 0
B. `a+b`
C. b
D. `b^(2)`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Define the function
Let \( y = \tan^{-1}(a + bx) \).

### Step 2: Differentiate the function
To find the derivative \( \frac{dy}{dx} \), we will use the chain rule. The derivative of \( \tan^{-1}(u) \) where \( u = a + bx \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

Here, \( u = a + bx \), so \( \frac{du}{dx} = b \).

Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{1 + (a + bx)^2} \cdot b \]

### Step 3: Substitute \( x = 0 \)
Now, we need to evaluate the derivative at \( x = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{b}{1 + (a + b \cdot 0)^2} = \frac{b}{1 + a^2} \]

### Step 4: Set the derivative equal to 1
According to the problem, this derivative equals 1 when \( x = 0 \):
\[ \frac{b}{1 + a^2} = 1 \]

### Step 5: Solve for \( 1 + a^2 \)
We can rearrange this equation:
\[ b = 1 + a^2 \]

### Conclusion
Thus, we find that:
\[ 1 + a^2 = b \]

### Final Answer
The value of \( 1 + a^2 \) is \( b \).

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