Q. If variables $x$ and $y$ are related by the equation $x = \int_{0}^{y}{\frac{1}{\sqrt{1 + 9u^{2}}}du,}$ then $\frac{d^{2}y}{dx^{2}}$ is equal to
A) $\sqrt{1 + 9y^{2}}$
B) $\frac{1}{1 + 9y^{2}}$
C) $9y$
D) $\frac{1}{9}y$
Explanation
(c)
We know that be Newton’s Leibnitz formula
If $I = \int_{u}^{v}{f(t)dx}$,
Then $\frac{dI}{dx} = f(v)\frac{dv}{dx} - f(u)\frac{du}{dx}$
Where $u$ and $v$ are function of $x$
$\therefore\frac{dx}{dy} = \frac{1}{\sqrt{1 + 9y^{2}}}$
$\Rightarrow \frac{dy}{dx} = \sqrt{1 + 9y^{2}}$
$\therefore\frac{d^{2}y}{dx^{2}} = \frac{9y}{\sqrt{1 + 9y^{2}}}.\frac{dy}{dx}$
$= \frac{9y}{\sqrt{1 + 9y^{2}}}\sqrt{1 + 9y^{2}} = 9y$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.