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

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(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(

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

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(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

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