Question

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

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

Answer 1

Answer: (A)

$x=\underset{0}{\overset{y}{\int }}\frac{dt}{\sqrt{1+t^2}}$
$\Rightarrow \frac{1}{\sqrt{1+y^2}}.\frac{dy}{dx}$
$\left[\because IfI\left(x\right)=\underset{\varphi \left(x\right)}{\overset{\psi \left(x\right)}{\int }}f\left(t\right)dt,then\frac{dI\left(x\right)}{dx}=f\left\{\psi \left(x\right)\right\}.\left\{\frac{d}{dx}\psi \left(x\right)\right\}-f\left\{\varphi \left(x\right)\right\}.\left\{\frac{d}{dx}\varphi \left(x\right)\right\}\right]$
$\frac{dy}{dx}=\sqrt{1-y^2}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{2\sqrt{1+y^2}}.2y.\frac{dy}{dx}=\frac{y}{\sqrt{1+y^2}}.\sqrt{1+y^2}=y$