The differential equation of y = sec(tan^{-1} | vedclass

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

Question

(C) Given the equation: $y = \sec(	an^{-1}x)$.Differentiating both sides with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}[\sec(

Vedclass · Accepted Answer

$(1 + x^2)\frac{dy}{dx} = xy$

Vedclass · Answer

(C) Given the equation: $y = \sec(	an^{-1}x)$.Differentiating both sides with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}[\sec(	an^{-1}x)]$Using the derivative formula $\frac{d}{dx}[\sec(u)] = \sec(u)	an(u)\frac{du}{dx}$ and $\frac{d}{dx}[	an^{-1}x] = \frac{1}{1+x^2}$:$\frac{dy}{dx} = \sec(	an^{-1}x) \cdot 	an(	an^{-1}x) \cdot \frac{1}{1+x^2}$Since $	an(	an^{-1}x) = x$ and $y = \sec(	an^{-1}x)$:$\frac{dy}{dx} = y \cdot x \cdot \frac{1}{1+x^2}$Rearranging the terms:$(1 + x^2)\frac{dy}{dx} = xy$.

The differential equation of $y = \sec(\tan^{-1}x)$ is

Similar Questions

The differential equation for which $\sqrt{1+y^2}=C x e^{\tan ^{-1} x}$ is the general solution,is

The differential equation of a family of hyperbolas whose axes are parallel to coordinate axes,centres lie on the line $y=2x$ and eccentricity is $\sqrt{3}$ is

The differential equation of the family of circles passing through $(0,0)$ and having centre on the $X$-axis is

$y = ax + b$ is

The differential equation of the family of curves $y = ax + \frac{1}{a}$,where $a (\neq 0)$ is an arbitrary constant,has the degree

Vedclass Test Series

Exam Paper Generator

Online Exam Module