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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $\frac{dy}{dx} =$

Similar Questions

For $x > 1$,if $(2x)^{2y} = 4e^{2x-2y}$,then $(1 + \log_e 2x)^2 \frac{dy}{dx}$ is equal to

If ${x^2}{e^y} + 2xy{e^x} + 13 = 0$,then $\frac{dy}{dx} = $

If $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$,then $\frac{dy}{dx}$ is equal to

If $\sin y = x \sin (a + y),$ then $\frac{dy}{dx} = $

If $\tan y = \cot \left(\frac{\pi}{4} - x\right)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module