Limits, Continuity & Differentiability Question - None

Question

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

Accepted Answer

Step 1: Given that $x^m y^n = (x+y)^{m+n}$

Step 2: Taking logarithm on both sides:

$m \log x + n \log y = (m+n) \log(x+y)$

Step 3: On differentiating w.r.t. $x$:

$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = (m+n)\frac{1}{x+y}\left(1 + \frac{dy}{dx}\right)$

Step 4: Rearranging:

$\frac{dy}{dx}\left[\frac{n}{y} - \frac{m+n}{x+y}\right] = \frac{m+n}{x+y} - \frac{m}{x}$

$\frac{dy}{dx}\left[\frac{n(x+y) - (m+n)y}{y(x+y)}\right] = \frac{m(x+y) - n(x+y) - mx}{x(x+y)}$

$\frac{dy}{dx}\left[\frac{nx - my}{y(x+y)}\right] = \frac{my - nx}{x(x+y)}$

Step 5: Therefore:

$\frac{dy}{dx} = \frac{y}{x}$

∴ Answer is (d).

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