Let `f(x)=sin^(-1)((2x+2)/(sqrt(4x^(2)+8x+13)))`, then the value of `(d(tanf(x)))/(d(tan^(-1)x))`, when `x=(1)/(2),` is..........

Answer

To solve the problem, we need to find the value of \(\frac{d(\tan(f(x)))}{d(\tan^{-1}(x))}\) when \(x = \frac{1}{2}\), given that \(f(x) = \sin^{-1}\left(\frac{2x + 2}{\sqrt{4x^2 + 8x + 13}}\right)\).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \sin^{-1}\left(\frac{2x + 2}{\sqrt{4x^2 + 8x + 13}}\right) \]

2. **Set \(f(x) = \theta\)**:
\[ \theta = f(x) \implies \sin(\theta) = \frac{2x + 2}{\sqrt{4x^2 + 8x + 13}} \]

3. **Use a right triangle to express \(\tan(\theta)\)**:
- Let the opposite side (perpendicular) be \(2x + 2\) and the hypotenuse be \(\sqrt{4x^2 + 8x + 13}\).
- By Pythagorean theorem, the base can be calculated:
\[ \text{hypotenuse}^2 = \text{perpendicular}^2 + \text{base}^2 \]
\[ 4x^2 + 8x + 13 = (2x + 2)^2 + b^2 \]
Expanding the right side:
\[ 4x^2 + 8x + 4 + b^2 = 4x^2 + 8x + 13 \]
Simplifying gives:
\[ b^2 = 9 \implies b = 3 \]

4. **Calculate \(\tan(\theta)\)**:
\[ \tan(\theta) = \frac{\text{perpendicular}}{\text{base}} = \frac{2x + 2}{3} \]

5. **Express \(f(x)\) in terms of \(\tan\)**:
\[ f(x) = \tan^{-1}\left(\frac{2x + 2}{3}\right) \]

6. **Differentiate \(\tan(f(x))\)**:
Using the chain rule:
\[ \frac{d}{dx}(\tan(f(x))) = \sec^2(f(x)) \cdot f'(x) \]

7. **Differentiate \(\tan^{-1}(x)\)**:
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]

8. **Combine results**:
\[ \frac{d(\tan(f(x)))}{d(\tan^{-1}(x))} = \frac{\sec^2(f(x)) \cdot f'(x)}{\frac{1}{1 + x^2}} = \sec^2(f(x)) \cdot f'(x) \cdot (1 + x^2) \]

9. **Calculate \(f'(x)\)**:
Using the derivative of \(\tan^{-1}\):
\[ f'(x) = \frac{2}{3(1 + x^2)} \]

10. **Evaluate at \(x = \frac{1}{2}\)**:
- First, calculate \(f\left(\frac{1}{2}\right)\):
\[ f\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{2} + 2}{3}\right) = \tan^{-1(1)} \]
Hence, \(\sec^2(f(\frac{1}{2})) = \sec^2\left(\frac{\pi}{4}\right) = 2\).

- Now, substitute \(x = \frac{1}{2}\) into \(f'(x)\):
\[ f'\left(\frac{1}{2}\right) = \frac{2}{3(1 + \left(\frac{1}{2}\right)^2)} = \frac{2}{3(1 + \frac{1}{4})} = \frac{2}{3 \cdot \frac{5}{4}} = \frac{8}{15} \]

11. **Final Calculation**:
\[ \frac{d(\tan(f(x)))}{d(\tan^{-1}(x))} = 2 \cdot \frac{8}{15} \cdot \left(1 + \left(\frac{1}{2}\right)^2\right) = 2 \cdot \frac{8}{15} \cdot \frac{5}{4} = \frac{80}{60} = \frac{4}{3} \]

### Final Answer:
The value of \(\frac{d(\tan(f(x)))}{d(\tan^{-1}(x))}\) when \(x = \frac{1}{2}\) is \(\frac{4}{3}\).