find the derivative of the function w.r.t `x sqrt((a^2-x^2)/(a^2+x^2))`

Answer

To find the derivative of the function \( y = \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the function
We can rewrite the function using exponent notation:
\[ y = \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{1/2} \]

### Step 2: Apply the Chain Rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The derivative of \( y \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{a^2 - x^2}{a^2 + x^2} \right) \]

### Step 3: Differentiate the inner function using the Quotient Rule
Let \( u = a^2 - x^2 \) and \( v = a^2 + x^2 \). We will apply the quotient rule:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\[ \frac{du}{dx} = -2x, \quad \frac{dv}{dx} = 2x \]
Now substituting back:
\[ \frac{d}{dx} \left( \frac{a^2 - x^2}{a^2 + x^2} \right) = \frac{(a^2 + x^2)(-2x) - (a^2 - x^2)(2x)}{(a^2 + x^2)^2} \]

### Step 4: Simplify the derivative
Now simplify the numerator:
\[ = \frac{-2x(a^2 + x^2) - 2x(a^2 - x^2)}{(a^2 + x^2)^2} \]
\[ = \frac{-2x(a^2 + x^2 + a^2 - x^2)}{(a^2 + x^2)^2} = \frac{-2x(2a^2)}{(a^2 + x^2)^2} = \frac{-4a^2x}{(a^2 + x^2)^2} \]

### Step 5: Substitute back into the derivative
Now substituting this back into the derivative expression:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{-1/2} \cdot \frac{-4a^2x}{(a^2 + x^2)^2} \]
\[ = \frac{-2a^2x}{(a^2 + x^2)^2 \sqrt{\frac{a^2 - x^2}{a^2 + x^2}}} \]

### Final Result
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = \frac{-2a^2x}{(a^2 + x^2)^2 \sqrt{\frac{a^2 - x^2}{a^2 + x^2}}} \]