On the curve `y= (1)/(1+ x^(2)),` the point at which `|(dy)/(dx )|` is greatest in the first quadrant is :

Options

A. `((1)/(2), (4)/(5))`
B. `(1, (1)/(4))`
C. `((1)/(sqrt2), (2)/(3))`
D. `((1)/( sqrt3), (3)/(4))`

Answer

To find the point on the curve \( y = \frac{1}{1 + x^2} \) where \( \left| \frac{dy}{dx} \right| \) is greatest in the first quadrant, we will follow these steps:

### Step 1: Find the derivative \( \frac{dy}{dx} \)

Given the function:
\[ y = \frac{1}{1 + x^2} \]

We can differentiate \( y \) with respect to \( x \) using the quotient rule or the chain rule. Here, we will use the chain rule:

\[ \frac{dy}{dx} = -\frac{d}{dx}(1 + x^2)^{-1} = -(-1)(1 + x^2)^{-2} \cdot (2x) = -\frac{2x}{(1 + x^2)^2} \]

### Step 2: Find the modulus of the derivative

We need to find \( \left| \frac{dy}{dx} \right| \):
\[ \left| \frac{dy}{dx} \right| = \frac{2x}{(1 + x^2)^2} \]

### Step 3: Maximize \( \left| \frac{dy}{dx} \right| \)

To find the maximum value of \( \left| \frac{dy}{dx} \right| \), we will differentiate \( \left| \frac{dy}{dx} \right| \) with respect to \( x \) and set it to zero.

Let:
\[ f(x) = \frac{2x}{(1 + x^2)^2} \]

Using the quotient rule to differentiate \( f(x) \):
\[ f'(x) = \frac{(1 + x^2)^2 \cdot 2 - 2x \cdot 2(1 + x^2)(2x)}{(1 + x^2)^4} \]

Simplifying the numerator:
\[ = 2(1 + x^2)^2 - 8x^2(1 + x^2) \]
\[ = 2(1 + 2x^2 + x^4 - 4x^2 - 4x^4) \]
\[ = 2(1 - 2x^2 - 3x^4) \]

Setting the numerator equal to zero:
\[ 1 - 2x^2 - 3x^4 = 0 \]

This is a quartic equation. Rearranging gives:
\[ 3x^4 + 2x^2 - 1 = 0 \]

Let \( u = x^2 \):
\[ 3u^2 + 2u - 1 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6} \]

Calculating the roots:
\[ u_1 = \frac{2}{6} = \frac{1}{3}, \quad u_2 = \frac{-6}{6} = -1 \text{ (not valid since } u = x^2 \text{ must be non-negative)} \]

Thus, \( x^2 = \frac{1}{3} \) implies:
\[ x = \frac{1}{\sqrt{3}} \]

### Step 5: Find the corresponding \( y \) value

Substituting \( x = \frac{1}{\sqrt{3}} \) into the equation for \( y \):
\[ y = \frac{1}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{1}{1 + \frac{1}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

### Conclusion

The point at which \( \left| \frac{dy}{dx} \right| \) is greatest in the first quadrant is:
\[ \left( \frac{1}{\sqrt{3}}, \frac{3}{4} \right) \]