`(d)/(dx)[tan^(-1)((12x-64x^(3))/(1-48x^(2)))]=`

Options

A. `(1)/(1+16x^(2))`
B. `(4)/(1+16x^(2))`
C. `(12)/(1+16x^(2))`
D. `(-12)/(1+16x^(2))`

Answer

To solve the problem \(\frac{d}{dx}\left[\tan^{-1}\left(\frac{12x - 64x^3}{1 - 48x^2}\right)\right]\), we will use the chain rule and the derivative of the inverse tangent function.

### Step 1: Identify the function
Let \( y = \tan^{-1}\left(\frac{12x - 64x^3}{1 - 48x^2}\right) \).

### Step 2: Differentiate using the chain rule
The derivative of \(\tan^{-1}(u)\) is given by:
\[ \frac{d}{dx}[\tan^{-1}(u)] = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{12x - 64x^3}{1 - 48x^2} \).

### Step 3: Find \( \frac{du}{dx} \)
To find \( \frac{du}{dx} \), we will use the quotient rule:
\[ \frac{du}{dx} = \frac{(1 - 48x^2)(12 - 192x^2) - (12x - 64x^3)(-96x)}{(1 - 48x^2)^2} \]
Calculating this step-by-step:

1. **Numerator**: Differentiate the numerator and denominator separately.
- The derivative of the numerator \( 12x - 64x^3 \) is \( 12 - 192x^2 \).
- The derivative of the denominator \( 1 - 48x^2 \) is \( -96x \).

2. **Apply the quotient rule**:
\[ \frac{du}{dx} = \frac{(1 - 48x^2)(12 - 192x^2) + (12x - 64x^3)(96x)}{(1 - 48x^2)^2} \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{12x - 64x^3}{1 - 48x^2}\right)^2} \cdot \frac{du}{dx} \]

### Step 5: Simplify the expression
To simplify \( 1 + u^2 \):
\[ 1 + \left(\frac{12x - 64x^3}{1 - 48x^2}\right)^2 = \frac{(1 - 48x^2)^2 + (12x - 64x^3)^2}{(1 - 48x^2)^2} \]

### Final expression
Thus, the final derivative is:
\[ \frac{dy}{dx} = \frac{(1 - 48x^2)^2 + (12x - 64x^3)^2}{(1 - 48x^2)^2} \cdot \frac{(1 - 48x^2)(12 - 192x^2) + (12x - 64x^3)(96x)}{(1 - 48x^2)^2} \]