Differentiation Question 52
Question: If $ x^{m}y^{n}={{(x+y)}^{m+n}} $ , then $ dy/dx $ is equal to
Options:
A) $ \frac{y}{x} $
B) $ \frac{x+y}{xy} $
C) $ xy $
D) $ \frac{x}{y} $
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Answer:
Correct Answer: A
Solution:
[a] $ x^{m}y^{n}={{(x+y)}^{m+n}} $
$ \therefore m\log x+n\log y=(m+n)In(x+y) $ Diffrentiating w.r.t.x,
$ \therefore \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}( 1+\frac{dy}{dx} ) $
$ \Rightarrow ( \frac{m}{x}-\frac{m+n}{x+y} )=( \frac{m+n}{x+y}-\frac{n}{y} )\frac{dy}{dx} $
$ \Rightarrow \frac{my-nx}{x(x+y)}=( \frac{my-nx}{y(x+y)} )\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{y}{x} $
BETA
