A) \[\frac{dy}{dx}=\frac{1+{{y}^{2}}}{1+{{x}^{2}}}\]
B) \[\frac{dy}{dx}=\frac{1+{{x}^{2}}}{1+{{y}^{2}}}\]
C) \[(1+{{x}^{2}})dy+(1+{{y}^{2}})dx=0\]
D) \[\frac{dy}{dx}=\frac{1-{{y}^{2}}}{1-{{x}^{2}}}\]
E) \[(1-{{x}^{2}})dx+(1-y)dy=0\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}x+{{\tan }^{-1}}y=c\] On differentiating w.r.t\[x\]we get \[\frac{1}{1+{{x}^{2}}}+\frac{1}{1+{{y}^{2}}}\frac{dy}{dx}=0\] \[\Rightarrow \] \[(1+{{x}^{2}})dy(1+{{y}^{2}})dx=0\]
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