If x = int_0^y frac{1}{sqrt{1 + 9t^2}} dt and | vedclass

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

Question

(C) Given $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$. By the Fundamental Theorem of Calculus,differentiating both sides with respect to $y$,we get $\

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Given $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$. By the Fundamental Theorem of Calculus,differentiating both sides with respect to $y$,we get $\frac{dx}{dy} = \frac{1}{\sqrt{1 + 9y^2}}$. Taking the reciprocal,we have $\frac{dy}{dx} = \sqrt{1 + 9y^2}$. Differentiating both sides with respect to $x$ using the chain rule,we get $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sqrt{1 + 9y^2}) = \frac{1}{2\sqrt{1 + 9y^2}} \cdot (18y) \cdot \frac{dy}{dx}$. Substituting $\frac{dy}{dx} = \sqrt{1 + 9y^2}$,we get $\frac{d^2y}{dx^2} = \frac{9y}{\sqrt{1 + 9y^2}} \cdot \sqrt{1 + 9y^2} = 9y$. Comparing $\frac{d^2y}{dx^2} = 9y$ with $\frac{d^2y}{dx^2} = ay$,we find $a = 9$.

If $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$ and $\frac{d^2y}{dx^2} = ay$,then $a$ is equal to

Similar Questions

If $y = \sin(\log_e x)$,then $x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx}$ is equal to

If $f(x) = \sin x + \cos x$,then $f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right)$ is equal to

If $y = \sin x + e^x,$ then $\frac{d^2x}{dy^2} = $

If $y^2 = ax^2 + bx + c$,where $a, b, c$ are constants,then $y^3 \frac{d^2 y}{dx^2}$ is equal to

If $y=2 \cos (2 \log x)+3 \sin (2 \log x)$,then $x^2 y^{\prime \prime}+x y^{\prime}+2 y=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module