If `f'(x)=(1)/((1+x^(2))^(3//2))` and `f(0)=0,` then `f(1)` is equal to :

Options

A. `sqrt2`
B. `-1/(sqrt(2))`
C. `1/(sqrt(2))`
D. None of these

Explanation

To solve the problem step by step, we need to find the function \( f(x) \) given its derivative \( f'(x) \) and the initial condition \( f(0) = 0 \).

### Step 1: Integrate the derivative

Given:
\[ f'(x) = \frac{1}{(1 + x^2)^{3/2}} \]
We need to find \( f(x) \) by integrating \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int \frac{1}{(1 + x^2)^{3/2}} \, dx \]

### Step 2: Use a trigonometric substitution

To solve the integral, we can use the substitution \( x = \tan(\theta) \). Then, \( dx = \sec^2(\theta) \, d\theta \) and \( 1 + x^2 = \sec^2(\theta) \).

Substituting these into the integral gives:
\[ f(x) = \int \frac{\sec^2(\theta)}{(\sec^2(\theta))^{3/2}} \, d\theta = \int \frac{\sec^2(\theta)}{\sec^3(\theta)} \, d\theta = \int \cos(\theta) \, d\theta \]

### Step 3: Integrate

The integral of \( \cos(\theta) \) is:
\[ f(x) = \sin(\theta) + C \]

### Step 4: Substitute back to original variable

Since \( \theta = \tan^{-1}(x) \), we have:
\[ f(x) = \sin(\tan^{-1}(x)) + C \]

Using the right triangle definition, where \( \tan(\theta) = \frac{x}{1} \), we can find:
- Opposite side = \( x \)
- Adjacent side = \( 1 \)
- Hypotenuse = \( \sqrt{1 + x^2} \)

Thus,
\[ \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} \]
So,
\[ f(x) = \frac{x}{\sqrt{1 + x^2}} + C \]

### Step 5: Use the initial condition to find \( C \)

We know \( f(0) = 0 \):
\[ f(0) = \frac{0}{\sqrt{1 + 0^2}} + C = 0 \implies C = 0 \]
Thus,
\[ f(x) = \frac{x}{\sqrt{1 + x^2}} \]

### Step 6: Find \( f(1) \)

Now we can find \( f(1) \):
\[ f(1) = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Final Answer
\[ f(1) = \frac{\sqrt{2}}{2} \]