Q. Differential equation of $y = \sec{(\tan^{- 1}{x)}}$ is
A) $\left( 1 + x^{2} \right)\frac{dy}{dx} = y + x$
B) $\left( 1 + x^{2} \right)\frac{dy}{dx} = y - x$
C) $\left( 1 + x^{2} \right)\frac{dy}{dx} = xy$
D) $\left( 1 + x^{2} \right)\frac{dy}{dx} = \frac{x}{y}$
Explanation
(c)
Given equation is $y = \sec{(\tan^{- 1}{x)}}$
On differentiating $w.r.t.x$, we get
$\frac{dy}{dx} = \sec{(\tan^{- 1}{x)\tan{(\tan^{- 1}{x) \bullet \frac{1}{1 + x^{2}} = \frac{xy}{1 + x^{2}}}}}\ \ \lbrack\because\tan{(\tan^{- 1}{x) = x\rbrack}}}$
$\Rightarrow \left( 1 + x^{2} \right)\frac{dy}{dx} = xy$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.