A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
\[{{x}^{4}}+{{y}^{4}}={{\left( t-\frac{1}{t} \right)}^{2}}+2={{({{x}^{2}}+{{y}^{2}})}^{2}}+2\] Þ \[{{x}^{2}}{{y}^{2}}=-1\Rightarrow {{y}^{2}}=-\frac{1}{{{x}^{2}}}\] Differentiating, we get \[2y\frac{dy}{dx}=\frac{2}{{{x}^{3}}}\]or\[{{x}^{3}}y\frac{dy}{dx}=1\].You need to login to perform this action.
You will be redirected in
3 sec