Limits Continuity And Differentiability Question 29
Question: If $ x^{a}y^{b}={{(x-y)}^{a+b}}, $ then the value of $ \frac{dy}{dx}-\frac{y}{x} $ is equal to
Options:
A) $ \frac{a}{b} $
B) $ \frac{b}{a} $
C) 1
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
$ x^{a}y^{b}={{(x-y)}^{a+b}} $ Taking log both the sides $ \log ( x^{a}y^{b} )=\log {{(x-y)}^{(a+b)}} $
$ a\log x+b\log y=(a+b)\log (x-y) $ differentiating both sides w.r.t ?x?. $ \frac{a}{x}+\frac{b}{y}\frac{dy}{dx}=\frac{(a+b)}{(x-y)}[ 1-\frac{dy}{dx} ] $
$ \frac{dy}{dx}[ \frac{b}{y}+\frac{a+b}{x-y} ]=\frac{a+b}{x-y}-\frac{a}{x} $
$ \frac{dy}{dx}[ \frac{bx-by+ay+by}{y(x-y)} ]=\frac{ax+bx-ax+ay}{x(x-y)} $
$ \frac{dy}{dx}[ \frac{bx+ay}{y} ]=\frac{bx+ay}{x} $
$ \frac{dy}{dx}=\frac{y}{x};\frac{dy}{dx}-\frac{y}{x}=0 $
BETA
