Differentiate ` tan ^(-1) ((8x)/(1-15x^(2))) ` w.r.to x

Answer

To differentiate the function \( y = \tan^{-1} \left( \frac{8x}{1 - 15x^2} \right) \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the function
Let
\[ y = \tan^{-1} \left( \frac{8x}{1 - 15x^2} \right) \]

### Step 2: Use the formula for the tangent of a sum
We can express \( \frac{8x}{1 - 15x^2} \) as a sum of two terms:
\[ \frac{8x}{1 - 15x^2} = \frac{5x + 3x}{1 - 5x \cdot 3x} \]
This can be rewritten using the identity for the tangent of a sum:
\[ y = \tan^{-1}(5x) + \tan^{-1}(3x) \]

### Step 3: Differentiate both sides
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}(5x) \right) + \frac{d}{dx} \left( \tan^{-1}(3x) \right) \]

### Step 4: Apply the derivative of the inverse tangent function
The derivative of \( \tan^{-1}(u) \) is given by \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \). Therefore:
\[ \frac{dy}{dx} = \frac{1}{1 + (5x)^2} \cdot 5 + \frac{1}{1 + (3x)^2} \cdot 3 \]

### Step 5: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = \frac{5}{1 + 25x^2} + \frac{3}{1 + 9x^2} \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{5}{1 + 25x^2} + \frac{3}{1 + 9x^2} \]

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