If `d/(dx)(tan^(-1)(2^(x+1)/(1-4^(x))))=a^(x+1)/(1+b^(x))loga`, find the value of a and b.

Answer

To solve the problem, we will differentiate the given expression and then equate it to the right-hand side to find the values of \( a \) and \( b \).

### Step 1: Define the function
Let:
\[ u = \tan^{-1}\left(\frac{2^{x+1}}{1 - 4^x}\right) \]

### Step 2: Simplify the argument of the arctangent
Notice that:
\[ 4^x = (2^2)^x = (2^x)^2 \]
Thus, we can rewrite the expression:
\[ u = \tan^{-1}\left(\frac{2 \cdot 2^x}{1 - (2^x)^2}\right) \]

### Step 3: Use the double angle formula
Using the identity for the tangent of a double angle:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
we can set \( \tan(\theta) = 2^x \). Therefore:
\[ u = \tan^{-1}(\tan(2\theta)) = 2\theta \]

### Step 4: Express \( \theta \) in terms of \( x \)
Since \( \theta = \tan^{-1}(2^x) \), we have:
\[ u = 2 \tan^{-1}(2^x) \]

### Step 5: Differentiate \( u \)
Now we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(2^x)) \]
Using the chain rule:
\[ \frac{d}{dx}(\tan^{-1}(2^x)) = \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x) = \frac{1}{1 + 4^x} \cdot (2^x \ln(2)) \]
Thus,
\[ \frac{du}{dx} = 2 \cdot \frac{2^x \ln(2)}{1 + 4^x} \]

### Step 6: Set the derivative equal to the right-hand side
We have:
\[ \frac{du}{dx} = \frac{2^{x+1} \ln(2)}{1 + 4^x} \]
According to the problem, this is equal to:
\[ \frac{a^{x+1}}{1 + b^x \log a} \]

### Step 7: Compare both sides
From the expressions, we can equate:
\[ \frac{2^{x+1} \ln(2)}{1 + 4^x} = \frac{a^{x+1}}{1 + b^x \log a} \]

### Step 8: Identify \( a \) and \( b \)
To find \( a \) and \( b \), we can compare coefficients:
1. From the numerator, we see that \( a^{x+1} = 2^{x+1} \ln(2) \), which leads us to conclude \( a = 2 \) and \( \ln(2) = 1 \).
2. From the denominator, \( 1 + 4^x \) can be rewritten as \( 1 + (2^2)^x = 1 + (b^x \log a) \). This implies \( b = 4 \).

### Final Values
Thus, we find:
\[ a = 2, \quad b = 4 \]