Differentiation Question 301
Question: If $ x^{2}+y^{2}=t-\frac{1}{t}, $
$ x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}} $ , then $ x^{3}y\frac{dy}{dx}= $
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
$ x^{4}+y^{4}={{( t-\frac{1}{t} )}^{2}}+2={{(x^{2}+y^{2})}^{2}}+2 $
Therefore $ x^{2}y^{2}=-1\Rightarrow y^{2}=-\frac{1}{x^{2}} $
Differentiating, we get $ 2y\frac{dy}{dx}=\frac{2}{x^{3}} $ or $ x^{3}y\frac{dy}{dx}=1 $ .
BETA
