Question

If $x={\int }_0^y\frac{dt}{\sqrt{1+t^2}}$ then $\frac{d^{2}y}{d{x}^2}$ is:

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

Answer 1

The correct option is A $y$

x=${\int }_0^y\frac{dt}{\sqrt{1+t^2}}=ln(t+\sqrt{1+t^2}){|}_0^y=ln(y+\sqrt{1+y^2})$

$\to 1=\frac{1}{\sqrt{1+y^2}}\frac{dy}{dx}$

$\to \frac{dy}{dx}=\sqrt{1+y^2}$

$\to \frac{d^{2}y}{d{x}^2}=\frac{2y}{2\sqrt{1+y^2}}\frac{dy}{dx}=y$