If y sec x + tan x + x^2 y = 0,then frac{dy}{ | vedclass

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Question

(C) Given the equation: $y \sec x + 	an x + x^2 y = 0$. Differentiating both sides with respect to $x$ using the product rule and chain rule: $\frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the equation: $y \sec x + 	an x + x^2 y = 0$. Differentiating both sides with respect to $x$ using the product rule and chain rule: $\frac{d}{dx}(y \sec x) + \frac{d}{dx}(	an x) + \frac{d}{dx}(x^2 y) = 0$ $\left( \frac{dy}{dx} \sec x + y \sec x 	an x ight) + \sec^2 x + \left( 2xy + x^2 \frac{dy}{dx} ight) = 0$ Grouping the terms containing $\frac{dy}{dx}$: $\frac{dy}{dx} (\sec x + x^2) + y \sec x 	an x + \sec^2 x + 2xy = 0$ $\frac{dy}{dx} (\sec x + x^2) = -(2xy + \sec^2 x + y \sec x 	an x)$ $\frac{dy}{dx} = -\frac{2xy + \sec^2 x + y \sec x 	an x}{x^2 + \sec x}$.

If $y \sec x + \tan x + x^2 y = 0$,then $\frac{dy}{dx} =$

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