If `y = ((x^(a))/(x^(b)))^(a + b) ((x^(b))/(x^(c)))^(b + c) ((x^(c))/(x^(a)))^(c + a) , "then" (dy)/(dx) ` is equal to

Options

A. 0
B. ` a + b + c `
C. 1
D. None of these

Explanation

To solve the problem, we start with the given function:

\[ y = \left(\frac{x^a}{x^b}\right)^{a+b} \left(\frac{x^b}{x^c}\right)^{b+c} \left(\frac{x^c}{x^a}\right)^{c+a} \]

### Step 1: Simplify each term
We can simplify each fraction inside the parentheses:

1. \(\frac{x^a}{x^b} = x^{a-b}\)
2. \(\frac{x^b}{x^c} = x^{b-c}\)
3. \(\frac{x^c}{x^a} = x^{c-a}\)

Thus, we can rewrite \(y\) as:

\[ y = \left(x^{a-b}\right)^{a+b} \left(x^{b-c}\right)^{b+c} \left(x^{c-a}\right)^{c+a} \]

### Step 2: Apply the power rule
Using the power rule \((x^m)^n = x^{mn}\), we get:

\[ y = x^{(a-b)(a+b)} \cdot x^{(b-c)(b+c)} \cdot x^{(c-a)(c+a)} \]

### Step 3: Combine the exponents
Since the bases are the same, we can add the exponents:

\[ y = x^{(a-b)(a+b) + (b-c)(b+c) + (c-a)(c+a)} \]

### Step 4: Expand the exponents
Now we expand each term:

1. \((a-b)(a+b) = a^2 - b^2\)
2. \((b-c)(b+c) = b^2 - c^2\)
3. \((c-a)(c+a) = c^2 - a^2\)

Combining these gives:

\[ y = x^{(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)} \]

### Step 5: Simplify the exponent
Notice that when we combine the terms:

\[ y = x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2} \]

All the terms cancel out:

\[ y = x^0 \]

### Step 6: Final simplification
Since \(x^0 = 1\), we have:

\[ y = 1 \]

### Step 7: Differentiate \(y\)
Now, we differentiate \(y\) with respect to \(x\):

\[ \frac{dy}{dx} = 0 \]

Thus, the final answer is:

\[ \frac{dy}{dx} = 0 \]