find dy/dx
y sec x + tan x +x²y = 0
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Answer: y sec x + tan x + x2y = 0
differentiating both sides with respect to x
, we get⇒y . sec x tan x + sec x . dy/dx + sec^2x + x^2dydx + y . 2x = 0
⇒(sec x + x^2)dy/dx = −2xy − sec^2x − y sec x . tan x
⇒dy/dx = −(y sec x tan x + sec^2x + 2xy)/(sec x + x^2)
Step-by-step explanation:
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