Differentiation Question 327
Question: If $ x^{m}y^{n}={{(x+y)}^{m+n}} $ then $ {{. \frac{dy}{dx} |} _{x=1,y=2}} $ is equal to
[J & K 2005]
Options:
A) ½
B) 2
C) 2m/n
D) m/ 2n
Show Answer
Answer:
Correct Answer: B
Solution:
$ x^{m}y^{n}={{(x+y)}^{m+n}} $
$ x^{m}.(n{y^{n-1}}).\frac{dy}{dx}+y^{n}(m{x^{m-1}})=(m+n){{(x+y)}^{m+n-1}} $
$ ( 1+\frac{dy}{dx} ) $
After solving, we find $ \frac{dy}{dx}=\frac{y}{x} $ and $ {{. \frac{dy}{dx} |} _{x=1,y=2}}=\frac{2}{1}=2 $ .
BETA
