Question

If \( x^m y^n = (x+y)^{m+n} \), then \( \frac{dy}{dx} \) is
 

• A. \( \frac{x+y}{xy} \)
• B. \( xy \)
• C. \( \frac{x}{y} \) (Typo in OCR)
• D. \( \frac{y}{x} \)

Hint:

For any implicit function of the form \( x^m y^n = (x+y)^{m+n} \), the derivative \( \frac{dy}{dx} \) is always \( \frac{y}{x} \). This is a standard result worth memorizing for competitive exams. Logarithmic differentiation is the key to proving it.

Answer 1

The Correct Option is D

This is a standard problem that can be solved efficiently using logarithmic differentiation.
Given equation: \( x^m y^n = (x+y)^{m+n} \). Take the natural logarithm (ln) on both sides: \[ \ln(x^m y^n) = \ln((x+y)^{m+n}) \] Using logarithm properties: \[ \ln(x^m) + \ln(y^n) = (m+n) \ln(x+y) \] \[ m \ln(x) + n \ln(y) = (m+n) \ln(x+y) \] Now, differentiate both sides with respect to x, remembering to use the chain rule for terms involving y.
\[ m \cdot \frac{1}{x} + n \cdot \frac{1}{y} \cdot \frac{dy}{dx} = (m+n) \cdot \frac{1}{x+y} \cdot \left(1 + \frac{dy}{dx}\right) \] \[ \frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} + \frac{m+n}{x+y} \frac{dy}{dx} \] Now, group the terms with \( \frac{dy}{dx} \) on one side and the other terms on the other side.
\[ \left(\frac{n}{y} - \frac{m+n}{x+y}\right) \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} \] Find common denominators for the terms in the parentheses: \[ \left(\frac{n(x+y) - y(m+n)}{y(x+y)}\right) \frac{dy}{dx} = \frac{x(m+n) - m(x+y)}{x(x+y)} \] \[ \left(\frac{nx + ny - my - ny}{y(x+y)}\right) \frac{dy}{dx} = \frac{mx + nx - mx - my}{x(x+y)} \] \[ \left(\frac{nx - my}{y(x+y)}\right) \frac{dy}{dx} = \frac{nx - my}{x(x+y)} \] Assuming \( nx - my \neq 0 \), we can cancel this term from both sides. We can also cancel \( (x+y) \). \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \] \[ \frac{dy}{dx} = \frac{y}{x} \]