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

Options

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

Answer

To solve the problem, we need to differentiate the function given by:

\[ y = \tan^{-1}\left(\frac{a+x}{1-ax}\right) \]

### Step 1: Differentiate using the chain rule

Using the derivative of the inverse tangent function, we have:

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

### Step 2: Differentiate the fraction

Now we need to differentiate the function \(\frac{a+x}{1-ax}\). We will use the quotient rule for differentiation, which states:

\[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \]

Here, \(f(x) = a+x\) and \(g(x) = 1-ax\).

- \(f'(x) = 1\)
- \(g'(x) = -a\)

Now applying the quotient rule:

\[ \frac{d}{dx}\left(\frac{a+x}{1-ax}\right) = \frac{(1-ax)(1) - (a+x)(-a)}{(1-ax)^2} \]

### Step 3: Simplify the numerator

Expanding the numerator:

\[ = \frac{(1-ax) + a(a+x)}{(1-ax)^2} \]
\[ = \frac{1 - ax + a^2 + ax}{(1-ax)^2} \]
\[ = \frac{1 + a^2}{(1-ax)^2} \]

### Step 4: Substitute back into the derivative

Now substituting this back into the derivative we found in Step 1:

\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{a+x}{1-ax}\right)^2} \cdot \frac{1 + a^2}{(1-ax)^2} \]

### Step 5: Simplify the expression

Next, we simplify \(\left(\frac{a+x}{1-ax}\right)^2\):

\[ \left(\frac{a+x}{1-ax}\right)^2 = \frac{(a+x)^2}{(1-ax)^2} \]

Thus,

\[ 1 + \left(\frac{a+x}{1-ax}\right)^2 = 1 + \frac{(a+x)^2}{(1-ax)^2} = \frac{(1-ax)^2 + (a+x)^2}{(1-ax)^2} \]

Now substituting this back into the derivative:

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

### Step 6: Final simplification

The \((1-ax)^2\) cancels out:

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

### Final Result

Thus, the final result for \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{1 + a^2}{1 + x^2 + a^2} \]