If ` y=sin ^(-1) ((2^(x+1))/( 1+4^(x))),then (dy)/(dx) =`

Options

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

Answer

To solve the problem, we need to find the derivative of the function \( y = \sin^{-1} \left( \frac{2^{x+1}}{1 + 4^x} \right) \).

### Step-by-Step Solution:

1. **Rewrite the Function:**
We can rewrite the expression inside the inverse sine function:
\[ y = \sin^{-1} \left( \frac{2^{x+1}}{1 + 4^x} \right) = \sin^{-1} \left( \frac{2 \cdot 2^x}{1 + (2^2)^x} \right) = \sin^{-1} \left( \frac{2 \cdot 2^x}{1 + 2^{2x}} \right) \]

2. **Use the Identity for Sine:**
We can use the identity \( \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)} \). Let \( \tan(\theta) = 2^x \):
\[ y = \sin^{-1} \left( \frac{2 \cdot 2^x}{1 + 2^{2x}} \right) = \sin^{-1}(\sin(2\theta)) = 2\theta \]
Therefore, we have:
\[ y = 2 \tan^{-1}(2^x) \]

3. **Differentiate Using the Chain Rule:**
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx} \left( \tan^{-1}(2^x) \right) \]
Using the derivative of \( \tan^{-1}(u) \), which is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \):
\[ \frac{d}{dx} \left( \tan^{-1}(2^x) \right) = \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x) \]

4. **Calculate the Derivative of \( 2^x \):**
The derivative of \( 2^x \) is:
\[ \frac{d}{dx}(2^x) = 2^x \ln(2) \]

5. **Combine the Results:**
Now substituting back:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + 2^{2x}} \cdot (2^x \ln(2)) \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{2 \cdot 2^x \ln(2)}{1 + 2^{2x}} \]

6. **Final Result:**
Therefore, the final result for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2^{x+1} \ln(2)}{1 + 4^x} \]