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If $y \sec x+\tan x+x^2 y=0$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$-\frac{2 x y+\sec ^2 x+y \sec x \tan x}{x^2+\sec x}$
$\begin{aligned} & y \sec x+\tan x+x^2 y=0 \\ & \Rightarrow \sec x \frac{d y}{d x}+y \sec x \tan x+\sec ^2 x+2 x y+x^2 \frac{d y}{d x}=0 \\ & \Rightarrow \frac{d y}{d x}=-\frac{2 x y+\sec ^2 x+y \sec x \tan x}{x^2+\sec x} .\end{aligned}$
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