`(d)/(dx)[tan^(-1)((ax-b)/(bx+a))]=`

Options

A. `(1)/(1+x^(2))+(a^(2))/(a^(2)+b^(2))`
B. `(-1)/(1+x^(2))-(a^(2))/(a^(2)+b^(2))`
C. `(1)/(1+x^(2))+(a^(2))/(a^(2)+b^(2))`
D. none of theses

Answer

To solve the problem \(\frac{d}{dx}\left[\tan^{-1}\left(\frac{ax - b}{bx + a}\right)\right]\), we will follow these steps:

### Step 1: Let \(y = \tan^{-1}\left(\frac{ax - b}{bx + a}\right)\)

This sets up our function for differentiation.

### Step 2: Let \(u = \frac{ax - b}{bx + a}\)

We will differentiate \(y\) with respect to \(x\) using the chain rule.

### Step 3: Differentiate \(y\) with respect to \(u\)

Using the derivative of the arctangent function:
\[ \frac{dy}{du} = \frac{1}{1 + u^2} \]

### Step 4: Differentiate \(u\) with respect to \(x\)

To find \(\frac{du}{dx}\), we will apply the quotient rule:
\[ u = \frac{f(x)}{g(x)} \quad \text{where } f(x) = ax - b \text{ and } g(x) = bx + a \]
The quotient rule states:
\[ \frac{du}{dx} = \frac{f' g - f g'}{g^2} \]
Calculating \(f'\) and \(g'\):
\[ f' = a, \quad g' = b \]
Thus,
\[ \frac{du}{dx} = \frac{(a)(bx + a) - (ax - b)(b)}{(bx + a)^2} \]

### Step 5: Simplify \(\frac{du}{dx}\)

Expanding the numerator:
\[ = \frac{abx + a^2 - abx + b^2}{(bx + a)^2} = \frac{a^2 + b^2}{(bx + a)^2} \]

### Step 6: Combine using the chain rule

Now, we can combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot \frac{a^2 + b^2}{(bx + a)^2} \]

### Step 7: Substitute \(u\) back into the equation

We need to substitute \(u = \frac{ax - b}{bx + a}\):
\[ 1 + u^2 = 1 + \left(\frac{ax - b}{bx + a}\right)^2 = \frac{(bx + a)^2 + (ax - b)^2}{(bx + a)^2} \]

### Step 8: Final expression for \(\frac{dy}{dx}\)

Thus, we have:
\[ \frac{dy}{dx} = \frac{(a^2 + b^2)}{(bx + a)^2} \cdot \frac{(bx + a)^2}{(bx + a)^2 + (ax - b)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{a^2 + b^2}{(bx + a)^2 + (ax - b)^2} \]

### Step 9: Conclusion

Thus, the final result is:
\[ \frac{d}{dx}\left[\tan^{-1}\left(\frac{ax - b}{bx + a}\right)\right] = \frac{a^2 + b^2}{(bx + a)^2 + (ax - b)^2} \]