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

Question

Aicharya · Accepted Answer

$x=\int_0^y \frac{d t}{\sqrt{1+t^2}}$ $\Rightarrow 1=\frac{1}{\sqrt{1+y^2}} \cdot \frac{d y}{d x}$ $\left[\because\right.$ If $\mathrm{I}(x)=\int_{\phi(x)}^{\psi(x)} f(t) d t$, then $\frac{d \mathrm{I}(x)}{d x}=f\{\psi(x)\}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{1-\mathrm{y}^2}$ $\frac{d^2 y}{d x^2}=\frac{1}{2 \sqrt{1+y^2}} \cdot 2 y \cdot \frac{d y}{d x}=\frac{y}{\sqrt{1+y^2}} \cdot \sqrt{1+y^2}=y$

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

Solution