If $ y \sec x+\tan x+x^{2} y=0 $, then $ \frac{d y}{d x} $ is equal to

$ -\frac{2 x y+\sec ^{2} x+\sec x \tan x}{x^{2}+\sec x} $

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

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Application +4 | -1

y \sec x+\tan x+x^{2} y=0

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the result of implicit differentiation.

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