First slide

Derivatives

Question

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

Moderate

A

$\frac{x + y}{xy}$
B

xy
C

$\frac{x}{y}$
D

$\frac{y}{x}$

Solution

Given that, $x^{m} y^{n} = (x + y)^{m + n}$
Taking on both sides, =we get
$mlog x + nlog y = (m + n) \log (x + y)$
On differentiating w.r.t. x, we get
$\begin{array}{l} \frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{(m + n)}{(x + y)} (1 + \frac{dy}{dx}) \\ \Rightarrow \frac{dy}{dx} (\frac{m + n}{x + y} - \frac{n}{y}) = \frac{m}{x} - \frac{m + n}{x + y} \\ \Rightarrow \frac{dy}{dx} (\frac{my + ny - nx - ny}{y (x + y)}) = \frac{mx + my - mx - nx}{x (x + y)} \\ ∴ \frac{dy}{dx} = \frac{y}{x} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App