If x = _0^y dt√1+t^2, then d^2ydx^2 is equal to:

Question

If $x = \displaystyle\int_0^y \dfrac{dt}{\sqrt{1+t^2}}$, then $\dfrac{d^2y}{dx^2}$ is equal to:

Accepted Answer

$x = \int_0^y \dfrac{dt}{\sqrt{1+t^2}} \Rightarrow 1 = \dfrac{1}{\sqrt{1+y^2}}\cdot\dfrac{dy}{dx}$

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

$\dfrac{d^2y}{dx^2} = \dfrac{1}{2\sqrt{1+y^2}}\cdot2y\cdot\dfrac{dy}{dx} = \dfrac{y}{\sqrt{1+y^2}}\cdot\sqrt{1+y^2} = y$

Solutions · 1