If xᵐ yⁿ = (x + y)ᵐ⁺ⁿ, then dy/dx is equal to

The correct answer is y/x.

To find the derivative $\frac{dy}{dx}$, we start with the given implicit relation:
$$x^m{y}^n={(x+y)}^{m+n}$$

Step 1: Apply logarithms to both sides
Taking the natural logarithm ($\ln$) of both sides simplifies the exponents:
$$\ln \left(x^m{y}^n\right)=\ln \left[{(x+y)}^{m+n}\right]$$

Using the logarithmic properties $\ln (A\cdot B)=\ln \left(A\right)+\ln \left(B\right)$ and $\ln \left(A^k\right)=k\ln \left(A\right)$, we rewrite the equation as:
$$m\ln \left(x\right)+n\ln \left(y\right)=(m+n)\ln (x+y)$$

Step 2: Differentiate with respect to $x$
Differentiating both sides with respect to $x$ and applying the chain rule:
$$\frac{m}{x}+\frac{n}{y}\cdot \frac{dy}{dx}=\frac{m+n}{x+y}\cdot \frac{d}{dx}(x+y)$$
Since $\frac{d}{dx}(x+y)=1+\frac{dy}{dx}$, we have:
$$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}(1+\frac{dy}{dx})$$

Step 3: Group terms with $\frac{dy}{dx}$
Expanding the right-hand side:
$$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}\frac{dy}{dx}$$
Rearranging the terms to group all $\frac{dy}{dx}$ terms on the left side:
$$(\frac{n}{y}-\frac{m+n}{x+y})\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}$$

Step 4: Simplify the expressions on both sides
Let us find a common denominator for the terms inside the parentheses on the left side:
$$\frac{n}{y}-\frac{m+n}{x+y}$$
Similarly, find a common denominator for the terms on the right side:
$$\frac{m+n}{x+y}$$

Substitute these simplified coefficients back into the equation:
$$\left[\frac{nx-my}{y(x+y)}\right]\frac{dy}{dx}=\frac{nx-my}{x(x+y)}$$

Step 5: Solve for $\frac{dy}{dx}$
By cancelling out the common factor $\frac{nx-my}{x+y}$ from both sides, we get:
$$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}$$
Multiplying both sides by $y$ yields the final result:
$$\frac{dy}{dx}=\frac{y}{x}$$