Find dy/dx, when: y sec x + tan x + x2 y = 0 The answer: -( y - Math - Continuity and Differentiability

Question

Find dy/dx, when:

y sec x + tan x + x2 y = 0

The answer: -( y sec x tanx + sec2 x + 2xy )/ (
x2 + sec x )

Varun.Rawat · Accepted Answer

We have, y sec x + tan x + x2y = 0differentiating both sides with
respect to x, we get⇒y sec x tan x + sec x dydx + sec2x +
x2dydx + y 2x = 0⇒sec x + x2dydx = -2xy - sec2x - y sec x tan
x⇒dydx = -y sec x tan x + sec2x + 2xysec x + x2

Find dy/dx, when: y sec x + tan x + x2 y = 0 The answer: -( y sec x tanx + sec2 x + 2xy )/ ( x2 + sec x )

Find dy/dx, when:

y sec x + tan x + x² y = 0

The answer: -( y sec x tanx + sec² x + 2xy )/ ( x² + sec x )