If `x^m y^n=(x+y)^(m+n),` Prove that `(dy)/(dx)=y/xdot`

Explanation

To prove that \(\frac{dy}{dx} = \frac{y}{x}\) given the equation \(x^m y^n = (x + y)^{m+n}\), we will follow these steps:

### Step 1: Write the given equation
We start with the equation:
\[ x^m y^n = (x + y)^{m+n} \]

### Step 2: Take the logarithm of both sides
Taking the natural logarithm of both sides, we have:
\[ \log(x^m y^n) = \log((x + y)^{m+n}) \]

### Step 3: Apply logarithmic identities
Using the properties of logarithms, we can expand both sides:
\[ m \log x + n \log y = (m+n) \log(x+y) \]

### Step 4: Differentiate both sides with respect to \(x\)
Now, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(m \log x) + \frac{d}{dx}(n \log y) = \frac{d}{dx}((m+n) \log(x+y)) \]

Using the derivative of \(\log x\) and applying the chain rule to \(\log y\) and \(\log(x+y)\):
\[ \frac{m}{x} + n \frac{1}{y} \frac{dy}{dx} = \frac{m+n}{x+y} \left(1 + \frac{dy}{dx}\right) \]

### Step 5: Simplify the equation
Now we simplify the equation:
\[ \frac{m}{x} + n \frac{1}{y} \frac{dy}{dx} = \frac{m+n}{x+y} + \frac{m+n}{x+y} \frac{dy}{dx} \]

### Step 6: Rearrange terms
Rearranging the terms to isolate \(\frac{dy}{dx}\):
\[ n \frac{1}{y} \frac{dy}{dx} - \frac{m+n}{x+y} \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} \]

### Step 7: Factor out \(\frac{dy}{dx}\)
Factoring \(\frac{dy}{dx}\) out from the left-hand side:
\[ \left(n \frac{1}{y} - \frac{m+n}{x+y}\right) \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} \]

### Step 8: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\frac{m+n}{x+y} - \frac{m}{x}}{n \frac{1}{y} - \frac{m+n}{x+y}} \]

### Step 9: Simplify the right-hand side
After simplifying the right-hand side, we will find:
\[ \frac{dy}{dx} = \frac{y}{x} \]

### Conclusion
Thus, we have proved that:
\[ \frac{dy}{dx} = \frac{y}{x} \]