If \[y\sec x+\tan x+{{x}^{2}}y=0\], then \[\frac{dy}{dx}\]= [DSSE 1981; CBSE 1981]

\[\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{{{x}^{2}}+\sec x}\]

\[-\frac{2xy+{{\sec }^{2}}x+\sec x\tan x}{{{x}^{2}}+\sec x}\]

\[-\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{{{x}^{2}}+\sec x}\]

If \[y\sec x+\tan x+{{x}^{2}}y=0\], then \[\frac{dy}{dx}\]= [DSSE 1981

1.	If \[y\sec x+\tan x+{{x}^{2}}y=0\], then \[\frac{dy}{dx}\]= [DSSE 1981; CBSE 1981]
A.	\[\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{{{x}^{2}}+\sec x}\]
B.	\[-\frac{2xy+{{\sec }^{2}}x+\sec x\tan x}{{{x}^{2}}+\sec x}\]
C.	\[-\frac{2xy+{{\sec }^{2}}x+y\sec x\tan x}{{{x}^{2}}+\sec x}\]
D.	None of these
Answer» D. None of these

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If \[y\sec x+\tan x+{{x}^{2}}y=0\], then \[\frac{dy}{dx}\]= [DSSE 1981; CBSE 1981]

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