Find the coordinates of the point on the curve, `y= (x^(2)-1)/(x^(2) + 1) (x gt 0)` where the gradient of the tangent to the curve is maximum

Answer

To find the coordinates of the point on the curve \( y = \frac{x^2 - 1}{x^2 + 1} \) (where \( x > 0 \)) where the gradient of the tangent to the curve is maximum, we can follow these steps:

### Step 1: Differentiate the function
We need to find the derivative of \( y \) with respect to \( x \) to determine the gradient of the tangent.

\[ y = \frac{x^2 - 1}{x^2 + 1} \]

Using the quotient rule for differentiation, which states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), we set \( u = x^2 - 1 \) and \( v = x^2 + 1 \).

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

\[ \frac{du}{dx} = 2x, \quad \frac{dv}{dx} = 2x \]

Now applying the quotient rule:

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

### Step 2: Simplify the derivative
Simplifying the numerator:

\[ \frac{dy}{dx} = \frac{2x(x^2 + 1 - (x^2 - 1))}{(x^2 + 1)^2} = \frac{2x(2)}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2} \]

### Step 3: Find the maximum gradient
To find where the gradient is maximum, we need to differentiate \( \frac{dy}{dx} \) again and set it to zero.

Let \( s = \frac{dy}{dx} = \frac{4x}{(x^2 + 1)^2} \).

Now, we differentiate \( s \):

Using the quotient rule again:

\[ \frac{ds}{dx} = \frac{(x^2 + 1)^2(4) - 4x(2)(x^2 + 1)(2x)}{(x^2 + 1)^4} \]

Simplifying the numerator:

\[ = \frac{4(x^2 + 1)^2 - 16x^2(x^2 + 1)}{(x^2 + 1)^4} \]

Setting the numerator to zero for critical points:

\[ 4(x^2 + 1)^2 - 16x^2(x^2 + 1) = 0 \]

Factoring out \( 4(x^2 + 1) \):

\[ 4(x^2 + 1)(x^2 + 1 - 4x^2) = 0 \]

This gives:

\[ x^2 + 1 - 4x^2 = 0 \implies -3x^2 + 1 = 0 \implies x^2 = \frac{1}{3} \implies x = \frac{1}{\sqrt{3}} \]

### Step 4: Find the corresponding \( y \) coordinate
Now substituting \( x = \frac{1}{\sqrt{3}} \) back into the original equation to find \( y \):

\[ y = \frac{\left(\frac{1}{\sqrt{3}}\right)^2 - 1}{\left(\frac{1}{\sqrt{3}}\right)^2 + 1} = \frac{\frac{1}{3} - 1}{\frac{1}{3} + 1} = \frac{-\frac{2}{3}}{\frac{4}{3}} = -\frac{1}{2} \]

### Step 5: Final coordinates
Thus, the coordinates of the point on the curve where the gradient of the tangent is maximum are:

\[ \left( \frac{1}{\sqrt{3}}, -\frac{1}{2} \right) \]