If `x^m y^n=(x+y)^(m+n),p rov et h a t(dy)/(dx)=y/xdot`

Explanation

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

### Step 1: Take the logarithm of both sides
We start with the equation:
\[ x^m y^n = (x+y)^{m+n} \]
Taking the logarithm of both sides gives:
\[ \log(x^m y^n) = \log((x+y)^{m+n}) \]

### Step 2: Apply logarithmic identities
Using the properties of logarithms, we can rewrite both sides:
\[ m \log x + n \log y = (m+n) \log(x+y) \]

### Step 3: Differentiate both sides with respect to \(x\)
Now, differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(m \log x + n \log y) = \frac{d}{dx}((m+n) \log(x+y)) \]
Using the chain rule and product rule, we get:
\[ \frac{m}{x} + n \frac{1}{y} \frac{dy}{dx} = (m+n) \left(\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right)\right) \]

### Step 4: Simplify the right-hand side
Expanding the right-hand side gives:
\[ \frac{m}{x} + n \frac{1}{y} \frac{dy}{dx} = \frac{(m+n)}{x+y} + \frac{(m+n)}{x+y} \frac{dy}{dx} \]

### Step 5: Rearranging terms
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ n \frac{1}{y} \frac{dy}{dx} - \frac{(m+n)}{x+y} \frac{dy}{dx} = \frac{m}{x} - \frac{(m+n)}{x+y} \]

### Step 6: Factor out \(\frac{dy}{dx}\)
Factoring out \(\frac{dy}{dx}\) from the left side:
\[ \frac{dy}{dx} \left(n \frac{1}{y} - \frac{(m+n)}{x+y}\right) = \frac{m}{x} - \frac{(m+n)}{x+y} \]

### Step 7: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\frac{m}{x} - \frac{(m+n)}{x+y}}{n \frac{1}{y} - \frac{(m+n)}{x+y}} \]

### Step 8: Simplify the expression
After simplifying the expression, we find that:
\[ \frac{dy}{dx} = \frac{y}{x} \]

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