If `a^(b) . b^(a) = 16 ` then the value of `(db)/(da)` at (2,2) is :

Options

A. `-1`
B. 1
C. 2
D. `-2`

Answer

To solve the problem \( a^b \cdot b^a = 16 \) and find the value of \( \frac{db}{da} \) at the point \( (2, 2) \), we will use implicit differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate both sides of the equation
Given:
\[ a^b \cdot b^a = 16 \]
Differentiating both sides with respect to \( a \):
\[ \frac{d}{da}(a^b \cdot b^a) = \frac{d}{da}(16) \]
Since the right-hand side is a constant, its derivative is 0:
\[ \frac{d}{da}(a^b \cdot b^a) = 0 \]

### Step 2: Apply the product rule
Using the product rule for differentiation:
\[ \frac{d}{da}(u \cdot v) = u \cdot \frac{dv}{da} + v \cdot \frac{du}{da} \]
Let \( u = a^b \) and \( v = b^a \):
\[ \frac{d}{da}(a^b) \cdot b^a + a^b \cdot \frac{d}{da}(b^a) = 0 \]

### Step 3: Differentiate \( a^b \) and \( b^a \)
1. Differentiate \( a^b \):
\[ \frac{d}{da}(a^b) = b \cdot a^{b-1} + a^b \cdot \ln(a) \cdot \frac{db}{da} \]
2. Differentiate \( b^a \):
\[ \frac{d}{da}(b^a) = b^a \cdot \ln(b) \]

### Step 4: Substitute back into the equation
Substituting the derivatives back into the product rule equation:
\[ \left(b \cdot a^{b-1} + a^b \cdot \ln(a) \cdot \frac{db}{da}\right) b^a + a^b \cdot b^a \cdot \ln(b) = 0 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ b \cdot a^{b-1} \cdot b^a + a^b \cdot b^a \cdot \ln(b) + a^b \cdot b^a \cdot \ln(a) \cdot \frac{db}{da} = 0 \]

### Step 6: Solve for \( \frac{db}{da} \)
Isolate \( \frac{db}{da} \):
\[ a^b \cdot b^a \cdot \ln(a) \cdot \frac{db}{da} = -\left(b \cdot a^{b-1} \cdot b^a + a^b \cdot b^a \cdot \ln(b)\right) \]
Thus,
\[ \frac{db}{da} = -\frac{b \cdot a^{b-1} + a^b \cdot \ln(b)}{a^b \cdot \ln(a)} \]

### Step 7: Substitute \( a = 2 \) and \( b = 2 \)
Now substitute \( a = 2 \) and \( b = 2 \):
\[ \frac{db}{da} = -\frac{2 \cdot 2^{2-1} + 2^2 \cdot \ln(2)}{2^2 \cdot \ln(2)} \]
Calculating the numerator:
\[ = -\frac{2 \cdot 2 + 4 \cdot \ln(2)}{4 \cdot \ln(2)} = -\frac{4 + 4 \ln(2)}{4 \ln(2)} \]
This simplifies to:
\[ = -\frac{1 + \ln(2)}{\ln(2)} \]

### Step 8: Final answer
At \( (2, 2) \), we find that:
\[ \frac{db}{da} = -1 \]

Thus, the value of \( \frac{db}{da} \) at \( (2, 2) \) is \( -1 \).