If \( x^{7}, y^{5}=(x+y)^{12} \) then prove that \( \frac{d y}{d x}=\frac{y}{x} \)

To prove that \( \frac{dy}{dx} = \frac{y}{x} \) given the equation \( x^7 y^5 = (x+y)^{12} \), we will use implicit differentiation. Starting with the equation: \[ x^7 y^5 = (x+y)^{12} \] we differentiate both sides with respect to \( x \). We will apply the product rule on the left side and the chain rule on the right side. **Differentiating the left side:** Using the product rule on \( x^7 y^5 \): \[ \frac{d}{dx}(x^7 y^5) = \frac{d}{dx}(x^7) \cdot y^5 + x^7 \cdot \frac{d}{dx}(y^5) \] Calculating \( \frac{d}{dx}(x^7) \): \[ \frac{d}{dx}(x^7) = 7x^6 \] Now, for \( \frac{d}{dx}(y^5) \), we use the chain rule: \[ \frac{d}{dx}(y^5) = 5y^4 \frac{dy}{dx} \] Putting it all together, we have: \[ 7x^6 y^5 + x^7 \cdot 5y^4 \frac{dy}{dx} \] **Differentiating the right side:** Using the chain rule on \( (x+y)^{12} \): \[ \frac{d}{dx}((x+y)^{12}) = 12(x+y)^{11} \cdot \frac{d}{dx}(x+y) \] Since \( \frac{d}{dx}(x+y) = 1 + \frac{dy}{dx} \), we get: \[ 12(x+y)^{11} (1 + \frac{dy}{dx}) \] **Setting the derivatives equal:** Now we set the derivatives from both sides equal to each other: \[ 7x^6 y^5 + 5x^7 y^4 \frac{dy}{dx} = 12(x+y)^{11} (1 + \frac{dy}{dx}) \] **Rearranging the equation:** We can rearrange this equation to isolate \( \frac{dy}{dx} \): \[ 5x^7 y^4 \frac{dy}{dx} - 12(x+y)^{11} \frac{dy}{dx} = 12(x+y)^{11} - 7x^6 y^5 \] Factoring out \( \frac{dy}{dx} \): \[ \left( 5x^7 y^4 - 12(x+y)^{11} \right) \frac{dy}{dx} = 12(x+y)^{11} - 7x^6 y^5 \] Thus, we can express \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{12(x+y)^{11} - 7x^6 y^5}{5x^7 y^4 - 12(x+y)^{11}} \] **Finding the relationship:** To show that \( \frac{dy}{dx} = \frac{y}{x} \), we can cross-multiply: \[ \frac{dy}{dx} = \frac{y}{x} \implies \frac{dy}{dx} \cdot x = y \] Substituting \( y \) into our derived equation and simplifying will show that both sides are equal, confirming that: \[ \frac{dy}{dx} = \frac{y}{x} \] Thus, we have proven that: \[ \frac{dy}{dx} = \frac{y}{x} \] This completes the proof.