Question

If $x= \int\limits^{y}_{0} \frac{dt}{\sqrt{1+t^{2}}}$, then $\frac{d^{2}y}{dx^{2}}$ is equal to:

• A. $y$
• B. $\sqrt{1+y^{2}}$
• C. $ \frac{x}{\sqrt{1+y^{2}}}$
• D. $y^2$

Answer 1

Answer: (A)

$x= \int\limits^{y}_{0} \frac{dt}{\sqrt{1+t^{2}}}$
$\Rightarrow \, \frac{1}{\sqrt{1+y^{2}}} . \frac{dy}{dx}$
$\left[\because \,If\, I\left(x\right) = \int\limits^{\psi\left(x\right)}_{\phi\left(x\right)} f\left(t\right) dt,\, then\, \frac{dI\left(x\right)}{dx} = f\left\{\psi\left(x\right)\right\}.\left\{\frac{d}{dx}\psi\left(x\right)\right\}-f\left\{\phi\left(x\right)\right\}. \left\{\frac{d}{dx}\phi\left(x\right)\right\}\right]$
$\frac{dy}{dx} = \sqrt{1-y^{2}}$
$\Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{1}{2\sqrt{1+y^{2}}}.2y. \frac{dy}{dx} = \frac{y}{\sqrt{1+y^{2}}}.\sqrt{1+y^{2}} = y$