Variables x and y are related by the equation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$.By the Fundamental Theorem of Calculus,differentiating both sides with respect to $x$ gives:

Vedclass · Accepted Answer

$y$

Vedclass · Answer

(B) Given $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$.By the Fundamental Theorem of Calculus,differentiating both sides with respect to $x$ gives:$\frac{dx}{dx} = \frac{d}{dx} \left( \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}} \right)$$1 = \frac{1}{\sqrt{1 + y^2}} \cdot \frac{dy}{dx}$Therefore,$\frac{dy}{dx} = \sqrt{1 + y^2}$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dx} (\sqrt{1 + y^2})$$\frac{d^2y}{dx^2} = \frac{1}{2\sqrt{1 + y^2}} \cdot 2y \cdot \frac{dy}{dx}$Substitute $\frac{dy}{dx} = \sqrt{1 + y^2}$ into the expression:$\frac{d^2y}{dx^2} = \frac{y}{\sqrt{1 + y^2}} \cdot \sqrt{1 + y^2} = y$.Thus,the correct option is $B$.

Variables $x$ and $y$ are related by the equation $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$. The value of $\frac{d^2y}{dx^2}$ is equal to

Similar Questions

If $f(x) = \frac{x^2}{x+a}$,then $f^{\prime \prime}(a)$ is equal to

If $f(x) = a\sin (\log x)$,then ${x^2}f''(x) + xf'(x) = . . . $

If $y=x+\tan x$,then $\cos ^2 x \frac{d^2 y}{d x^2}+2 x$ is equal to

If $y = x^3 \log(\log_e(1 + x))$,then $y''(0)$ equals

Find the value of $\frac{d^2x}{dy^2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module