If ` y=tan ^(-1) ((6x )/( 1-8x^(2))),then (dy)/(dx)=`

Options

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

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \tan^{-1}\left(\frac{6x}{1 - 8x^2}\right)\), we can use the properties of the inverse tangent function and differentiation. Here's a step-by-step solution:

### Step 1: Rewrite the function using the tangent addition formula
We can express \(y\) as:
\[ y = \tan^{-1}(6x) - \tan^{-1}(8x^2) \]
This is based on the identity for the tangent of the difference of two angles:
\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A - B}{1 + AB}\right) \]
In our case, we can set \(A = 6x\) and \(B = 8x^2\).

### Step 2: Differentiate using the derivative of the inverse tangent function
The derivative of \(\tan^{-1}(u)\) is given by:
\[ \frac{d}{dx}(\tan^{-1}(u)) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
We will apply this to both terms.

### Step 3: Differentiate \(y\)
1. For \(u = 6x\):
\[ \frac{du}{dx} = 6 \]
Therefore,
\[ \frac{d}{dx}(\tan^{-1}(6x)) = \frac{1}{1 + (6x)^2} \cdot 6 = \frac{6}{1 + 36x^2} \]

2. For \(v = 8x^2\):
\[ \frac{dv}{dx} = 16x \]
Therefore,
\[ \frac{d}{dx}(\tan^{-1}(8x^2)) = \frac{1}{1 + (8x^2)^2} \cdot 16x = \frac{16x}{1 + 64x^4} \]

### Step 4: Combine the derivatives
Now, we can combine the derivatives:
\[ \frac{dy}{dx} = \frac{6}{1 + 36x^2} - \frac{16x}{1 + 64x^4} \]

### Step 5: Simplify the expression
To combine these fractions, we need a common denominator:
\[ \frac{dy}{dx} = \frac{6(1 + 64x^4) - 16x(1 + 36x^2)}{(1 + 36x^2)(1 + 64x^4)} \]
Expanding the numerator:
\[ = \frac{6 + 384x^4 - 16x - 576x^3}{(1 + 36x^2)(1 + 64x^4)} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{384x^4 - 576x^3 - 16x + 6}{(1 + 36x^2)(1 + 64x^4)} \]

### Final Result
The derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{384x^4 - 576x^3 - 16x + 6}{(1 + 36x^2)(1 + 64x^4)} \]