`(d)/(dx)[sin^(-1)((2^(x+1))/(1+4^(x)))]`

Options

A. `(1)/(sqrt(1-2^(x)))`
B. `(log)/(sqrt(1-4^(x)))`
C. `(2^(x+1)log2)/(1+4^(x))`
D. `(log2)/(sqrt(1-2^(x)))`

Answer

To differentiate the function \( y = \sin^{-1}\left(\frac{2^{x+1}}{1 + 4^x}\right) \), we can follow these steps:

### Step 1: Simplify the Argument
First, we simplify the argument of the inverse sine function:
\[ \frac{2^{x+1}}{1 + 4^x} = \frac{2 \cdot 2^x}{1 + (2^2)^x} = \frac{2 \cdot 2^x}{1 + 2^{2x}}. \]

### Step 2: Use the Identity for Sine
Next, we can express the fraction in terms of tangent:
\[ \frac{2 \cdot 2^x}{1 + 2^{2x}} = \frac{2 \tan(\theta)}{1 + \tan^2(\theta)} = \sin(2\theta), \]
where we let \( 2^x = \tan(\theta) \). Thus, we can rewrite our function as:
\[ y = \sin^{-1}(\sin(2\theta)). \]

### Step 3: Simplify the Inverse Sine
Since \( \sin^{-1}(\sin(2\theta)) = 2\theta \) (for \( \theta \) in the appropriate range), we have:
\[ y = 2\theta. \]

### Step 4: Find \( \theta \) in Terms of \( x \)
Recall that \( \theta = \tan^{-1}(2^x) \). Therefore:
\[ y = 2 \tan^{-1}(2^x). \]

### Step 5: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(2^x)). \]

Using the derivative of \( \tan^{-1}(u) \), which is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \), we have:
\[ \frac{d}{dx}(\tan^{-1}(2^x)) = \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x). \]

### Step 6: Differentiate \( 2^x \)
The derivative of \( 2^x \) is:
\[ \frac{d}{dx}(2^x) = 2^x \ln(2). \]

### Step 7: Combine the Results
Now substituting back, we get:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + 4^x} \cdot (2^x \ln(2)). \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{2 \cdot 2^x \ln(2)}{1 + 4^x}. \]