Applications of Derivatives Question - None

Question

If $x^p \cdot y^q = (x + y)^{p+q}$, then $\frac{dy}{dx}$ is

Accepted Answer

Solution:

Given: $x^p \cdot y^q = (x + y)^{p+q}$

Taking logarithm on both sides:

$p\log x + q\log y = (p+q)\log(x+y)$

Differentiating with respect to $x$:

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

$\frac{p}{x} + \frac{q}{y}\frac{dy}{dx} = \frac{p+q}{x+y} + \frac{p+q}{x+y}\frac{dy}{dx}$

$\frac{q}{y}\frac{dy}{dx} - \frac{p+q}{x+y}\frac{dy}{dx} = \frac{p+q}{x+y} - \frac{p}{x}$

$\frac{dy}{dx}\left(\frac{q}{y} - \frac{p+q}{x+y}\right) = \frac{p+q}{x+y} - \frac{p}{x}$

After simplification:

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

This is independent of both $p$ and $q$.

Question: