Differentiation Question 447
Question: If $ x^{p}y^{q}={{(x+y)}^{p+q}}, $ then $ \frac{dy}{dx}= $
[RPET 1999; UPSEAT 2001]
Options:
A) $ \frac{y}{x} $
B) $ -\frac{y}{x} $
C) $ \frac{x}{y} $
D) $ -\frac{x}{y} $
Show Answer
Answer:
Correct Answer: A
Solution:
Taking $ \log $ both sides, $ p\log x+q\log y=(p+q)\log (x+y) $
Therefore $ \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{p+q}{x+y}( 1+\frac{dy}{dx} )\Rightarrow \frac{dy}{dx}=\frac{y}{x} $ .
BETA
