If `y=tan ^(-1) sqrt(( a-x)/( a+x)) ,then (dy)/(dx) =`

Options

A. ` (-1)/(2sqrt (a^(2)-x^(2))) `
B. ` (1)/(2sqrt (a^(2)-x^(2))) `
C. ` (-1)/(sqrt (a^(2)-x^(2))) `
D. ` (1)/(sqrt (a^(2)-x^(2))) `

Answer

To find the derivative of the function \( y = \tan^{-1} \left( \sqrt{\frac{a-x}{a+x}} \right) \), we will follow a step-by-step approach.

### Step 1: Identify the derivative of \( \tan^{-1}(x) \)
The derivative of \( y = \tan^{-1}(u) \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \sqrt{\frac{a-x}{a+x}} \).

**Hint:** Remember the derivative of the inverse tangent function and how to apply the chain rule.

### Step 2: Differentiate \( u = \sqrt{\frac{a-x}{a+x}} \)
To differentiate \( u \), we will first rewrite it as:
\[ u = \left( \frac{a-x}{a+x} \right)^{1/2} \]
Using the chain rule, we have:
\[ \frac{du}{dx} = \frac{1}{2} \left( \frac{a-x}{a+x} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{a-x}{a+x} \right) \]

**Hint:** Use the quotient rule to differentiate the fraction \( \frac{a-x}{a+x} \).

### Step 3: Differentiate \( \frac{a-x}{a+x} \) using the quotient rule
Let \( v = a - x \) and \( w = a + x \). Then, using the quotient rule:
\[ \frac{d}{dx} \left( \frac{v}{w} \right) = \frac{w \frac{dv}{dx} - v \frac{dw}{dx}}{w^2} \]
where \( \frac{dv}{dx} = -1 \) and \( \frac{dw}{dx} = 1 \). Therefore:
\[ \frac{d}{dx} \left( \frac{a-x}{a+x} \right) = \frac{(a+x)(-1) - (a-x)(1)}{(a+x)^2} \]
Simplifying this gives:
\[ \frac{-(a+x) - (a-x)}{(a+x)^2} = \frac{-2a}{(a+x)^2} \]

**Hint:** Simplify the expression carefully to avoid mistakes.

### Step 4: Substitute back to find \( \frac{du}{dx} \)
Now substituting back into the expression for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{1}{2} \left( \frac{a-x}{a+x} \right)^{-1/2} \cdot \frac{-2a}{(a+x)^2} \]
This simplifies to:
\[ \frac{du}{dx} = -\frac{a}{(a+x)^2 \sqrt{\frac{a-x}{a+x}}} \]

**Hint:** Ensure to keep track of the negative sign and the square root.

### Step 5: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative of \( y \)
Now substituting \( u \) and \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{a-x}{a+x} \right)} \cdot \left(-\frac{a}{(a+x)^2 \sqrt{\frac{a-x}{a+x}}}\right) \]
Simplifying \( 1 + \frac{a-x}{a+x} \) gives:
\[ \frac{(a+x) + (a-x)}{a+x} = \frac{2a}{a+x} \]
Thus:
\[ \frac{dy}{dx} = \frac{1}{\frac{2a}{a+x}} \cdot \left(-\frac{a}{(a+x)^2 \sqrt{\frac{a-x}{a+x}}}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{(a+x)}{2a} \cdot \frac{a}{(a+x)^2 \sqrt{\frac{a-x}{a+x}}} \]
Cancelling \( a+x \):
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{\frac{a-x}{a+x}}} \]

**Final Result:**
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{\frac{a-x}{a+x}}} \]